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Merge pull request #39 from AAluoxiang/master
第一周作业#1
2 parents 7fde135 + f738939 commit 1334f7a

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‎Week_01/id_1/LeetCode_1021_1.java

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package week01;
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import java.util.Stack;
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/**
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* @创建人 luoxiang
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* @创建时间 2019/6/6 18:00
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* @描述 LeetCode : 1021. 删除最外层的括号 https://leetcode-cn.com/problems/remove-outermost-parentheses/
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*/
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public class RemoveOuterMostParenteses {
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public static void main(String[] args) {
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String str = "(()())(())";
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String str2 = "(()())(())(()(()))";
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String str3 = "()()";
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System.out.println(new RemoveOuterMostParenteses().removeOuterParentheses2(str));
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System.out.println(new RemoveOuterMostParenteses().removeOuterParentheses2(str2));
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System.out.println(new RemoveOuterMostParenteses().removeOuterParentheses2(str3));
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}
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/**
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* Method 1 ： 使用 stack 来进行操作 。
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* ( 先判空,非空 则加入字符串中，在push;
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* ) 先pop,再判空，非空则加入 字符串中;
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* 时间复杂度 ： O(N) ; 空间复杂度 ： O(N)
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*/
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public String removeOuterParentheses(String S) {
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final Stack<Character> stack = new Stack<>();
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StringBuilder sb = new StringBuilder();
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for (char c : S.toCharArray()) {
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if (c == '(') {
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if (!stack.empty()) sb.append(c);
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stack.push(c);
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} else {
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stack.pop();
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if (!stack.empty()) sb.append(c);
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}
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}
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return sb.toString();
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}
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/**
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* Method 2 : 用一个计数器 count 来保存； （ count++ ; ) count-- ;
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* 时间复杂度 ： O(N) ; 空间复杂度 : O(1)
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*/
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public String removeOuterParentheses2(String S) {
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StringBuilder sb = new StringBuilder();
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int count = 0;
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for (char c : S.toCharArray()) {
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if (c == '(') {
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count++;
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if (count > 1) sb.append(c);
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} else {
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count--;
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if (count != 0) sb.append(c);
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}
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}
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return sb.toString();
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}
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}

‎Week_01/id_1/LeetCode_1047_1.java

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package week01;
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import java.util.Stack;
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/**
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* @创建人 luoxiang
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* @创建时间 2019/6/6 12:41
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* @描述 LeetCode : 1047. 删除字符串中的所有相邻重复项 https://leetcode-cn.com/problems/remove-all-adjacent-duplicates-in-string/
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*/
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public class RemoveAllAdjacent {
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public static void main(String[] args) {
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final RemoveAllAdjacent adjacent = new RemoveAllAdjacent();
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System.out.println(adjacent.removeDuplicates2("abbaca"));
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}
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/**
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* Method 1 : 使用堆栈 stack ;
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* 时间复杂度 ： O(N*K*K) ; 空间复杂度：O(N)
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*/
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public String removeDuplicates(String S) {
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final Stack<Character> stack = new Stack<>();
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for (Character item : S.toCharArray()) {
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if (!stack.empty() && stack.peek() == item) {
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stack.pop();
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} else {
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stack.push(item);
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}
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}
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final StringBuilder builder = new StringBuilder();
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while (!stack.empty()) {
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builder.append(stack.pop());
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}
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return builder.reverse().toString();
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}
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/**
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* Method 2 : 数组使用
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* 时间复杂度： O(S.length()) —> O(N) ； 空间复杂度： O(S.length) —> O(N)
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*/
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public String removeDuplicates2(String S) {
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final char[] chars = S.toCharArray();
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int top = 0;
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for (char c : chars) {
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if (top > 0 && chars[top - 1] == c) {
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top--;
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continue;
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}
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chars[top++] = c;
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}
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return new String(chars, 0, top);
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}
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}

‎Week_01/id_1/LeetCode_15_1.java

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package week01;
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import java.util.*;
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/**
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* @创建人 luoxiang
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* @创建时间 2019/6/5 16:34
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* @描述 LeerCode : 15. 三数之和 ; https://leetcode-cn.com/problems/3sum/
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*/
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public class ThreeSum {
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public static void main(String[] args) {
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final ThreeSum threeSum = new ThreeSum();
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int[] ints = {-1, 0, 1};
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int[] int2 = {-1, 0, 1, 2, -1, -4};
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int[] int3 = {0, 0, 0, 0};
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int[] int4 = {-4, -2, 1, -5, -4, -4, 4, -2, 0, 4, 0, -2, 3, 1, -5, 0};
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final List<List<Integer>> lists = threeSum.threeSum(ints);
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for (List<Integer> list : lists) {
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for (Integer integer : list) {
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System.out.print(integer + ",");
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}
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System.out.println();
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}
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System.out.println("-----------------------");
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final List<List<Integer>> lists2 = threeSum.threeSum(int2);
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for (List<Integer> list : lists2) {
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for (Integer integer : list) {
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System.out.print(integer + ",");
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}
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System.out.println();
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}
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System.out.println("-----------------------");
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final List<List<Integer>> lists3 = threeSum.threeSum(int3);
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for (List<Integer> list : lists3) {
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for (Integer integer : list) {
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System.out.print(integer + ",");
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}
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System.out.println();
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}
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System.out.println("-----------------------");
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final List<List<Integer>> lists4 = threeSum.threeSum(int4);
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for (List<Integer> list : lists4) {
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for (Integer integer : list) {
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System.out.print(integer + ",");
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}
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System.out.println();
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}
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System.out.println("----------------33333-------");
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final List<List<Integer>> lists5 = threeSum.threeSum3(int4);
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for (List<Integer> list : lists5) {
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for (Integer integer : list) {
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System.out.print(integer + ",");
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}
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System.out.println();
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}
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}
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/**
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* Method 1: 暴力法; 3 层循环 ——> 超时了......
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* 时间复杂度 ： O(N*N*N+NlogN) ; 空间复杂度 : O(N)
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*/
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public List<List<Integer>> threeSum(int[] nums) {
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Arrays.sort(nums);
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List<List<Integer>> lists = new LinkedList<>();
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for (int i = 0; i < nums.length - 2 && nums[i] <= 0; i++) {
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if (i == 0 || i > 0 && nums[i] != nums[i - 1]) {
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for (int j = i + 1; j < nums.length - 1 && nums[i] + nums[j] <= 0; j++) {
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if ( nums[j] != nums[j + 1]) {
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for (int k = j + 1; k < nums.length; k++) {
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// if (k > 2 && nums[k] == nums[k - 1]) continue;
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if (nums[i] + nums[j] + nums[k] == 0) {
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List<Integer> list = Arrays.asList(nums[i], nums[j], nums[k]);
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lists.add(list);
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}
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}
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}
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}
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}
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}
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return lists;
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}
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/**
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* 寻找新的解决方法
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* Method 3 : 排序后 ，再循环里面 ，从两边往中间 查找
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*/
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public List<List<Integer>> threeSum3(int[] nums) {
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Arrays.sort(nums);
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List<List<Integer>> lists = new LinkedList<>();
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for (int i = 0; i < nums.length - 2; i++) {
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if (i == 0 || (i > 0 && nums[i] != nums[i - 1])) {
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int left = i + 1, right = nums.length - 1, needValue = 0 - nums[i];
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while (left < right) {
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if (needValue == (nums[left] + nums[right])) {
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if (!lists.contains(Arrays.asList(nums[i], nums[left], nums[right]))) {
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lists.add(Arrays.asList(nums[i], nums[left], nums[right]));
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}
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while (left < right && nums[right] == nums[right - 1]) right--;
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while (left < right && nums[left] == nums[left + 1]) left++;
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right--;
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left++;
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} else if (needValue < (nums[left] + nums[right])) {
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right--;
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} else {
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left++;
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}
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}
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}
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}
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return lists;
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}
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}

‎Week_01/id_1/LeetCode_189_1.java

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package week01;
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/**
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* @创建人 luoxiang
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* @创建时间 2019/6/4 9:41
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* @描述 LeetCode:189. 旋转数组 https://leetcode-cn.com/problems/rotate-array/
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*/
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public class RotateArray {
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public static void main(String[] args) {
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final RotateArray rotateArray = new RotateArray();
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int[] nums = new int[]{-1, -100, 3, 99};
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rotateArray.rotate(nums, 2);
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for (int num : nums) {
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System.out.print(num + ",");
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}
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}
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/**
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* Function one :
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* k,两层循环， 第一层 数组 array; 第二层 k,要移动的位置
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* array[i] = array[length -1 ]
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* 时间复杂度 : O(k*n)
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* 空间复杂度 ： O(1)
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*/
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public static void rotate1(int[] nums, int k) {
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if (nums.length == 0 || k == 0) {
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return;
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}
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int n = nums.length;
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k %= n;
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for (int j = 1; j <= k; j++) {
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int tempLast = nums[n - 1];
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for (int i = n - 1; i > 0; i--) {
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nums[i] = nums[i - 1];
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}
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nums[0] = tempLast;
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}
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}
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/**
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* Function two:
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* 反转的思路:
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* 第一次： 全部反转 一遍
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* 接下来以 k 为节点 进行区分 reverse(nums,0,n-1)
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* 第二次： k 之前的 数据反转一遍 reverse(nums,0,k-1)
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* 第三次： k 之后的 数据也要反转一遍 reverse(nums,k,n-1)
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*
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* @param nums
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* @param k
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*/
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public void rotate(int[] nums, int k) {
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int n=nums.length;
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k %= n;
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reserve(nums,0,n-1);
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reserve(nums,0,k-1);
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reserve(nums,k,n-1);
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}
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private void reserve(int[] nums, int start, int end) {
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while (start < end) {
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int temp = nums[start];
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nums[start] = nums[end];
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nums[end] = temp;
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start++;
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end--;
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}
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}
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}

‎Week_01/id_1/LeetCode_1_1.java

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package week01;
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import java.util.HashMap;
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import java.util.Map;
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/**
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* @创建人 luoxiang
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* @创建时间 2019/6/6 10:20
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* @描述 LeetCode: 1. 两数之和 https://leetcode-cn.com/problems/two-sum/
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*/
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public class TwoNum {
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public static void main(String[] args) {
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final TwoNum twoNum = new TwoNum();
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int[] nums={3, 2, 4};
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final int[] ints = twoNum.twoSum2(nums, 6);
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for (int anInt : ints) {
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System.out.print(anInt+",");
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}
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}
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/**
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* Method 1 : 暴力法， 两层循环解决
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* 时间复杂度： O(N*N) ; 空间复杂度： O(N)
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*
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*/
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public int[] twoSum(int[] nums, int target) {
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for(int i=0;i<nums.length-1;i++){
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for(int j=i+1;j<nums.length;j++){
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if(target-nums[i]==nums[j]){
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return new int[]{i,j};
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}
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}
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}
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return new int[1];
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}
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/**
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* Method 2 : 使用Map来进行存储，key: 数组下标 ； value: 数组下标对应的值； 循环一遍，并把值存储到map中，后面的循环来进行比较；
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* 时间复杂度： O(N) ; 空间复杂度： O(N) ；
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*/
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public int[] twoSum2(int[] nums, int target) {
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final Map<Integer, Integer> map = new HashMap<>();
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for(int i=0;i<nums.length;i++){
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int temp=target-nums[i];
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if(map.containsKey(temp)){
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return new int[]{map.get(temp),i};
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}
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map.put(nums[i],i);
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}
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return new int[0];
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}
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}

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