Merge pull request #88 from PsycheYih/master

1st week leetcode solutions

Author: GeekUniversity

Week_01/id_46/leetcode_21_046.java

@@ -0,0 +1,25 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
+        if(l1 == null)
+            return l2;
+        if(l2 == null)
+            return l1;
+        if(l1.val >= l2.val){
+            l2.next = mergeTwoLists(l1, l2.next);
+            return l2;
+        }
+        else{
+            l1.next = mergeTwoLists(l1.next, l2);
+            return l1;
+        }
+    }
+
+}

Week_01/id_46/leetcode_24_046.java

@@ -0,0 +1,24 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode swapPairs(ListNode head) {
+    //递归
+    //终止条件
+        if(head == null || head.next == null)
+            return head;
+    //单元执行:交换head和head的下一个节点
+        ListNode next = head.next;
+        head.next = swapPairs(next.next);
+        next.next = head;
+    //返回值
+        return next;
+        
+    }
+    
+}

Week_01/id_46/leetcode_24_2_046.java

@@ -0,0 +1,24 @@
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    public ListNode swapPairs(ListNode head) {
+    //递归
+    //终止条件
+        if(head == null || head.next == null)
+            return head;
+    //单元执行:交换head和head的下一个节点
+        ListNode next = head.next;
+        head.next = swapPairs(next.next);
+        next.next = head;
+    //返回值
+        return next;
+        
+    }
+    
+}

Week_01/id_46/leetcode_26_046.java

@@ -0,0 +1,18 @@
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        if(nums.length == 0) return 0;
+        
+        int cur = 0;
+        for(int i = 0 ; i < nums.length ; i++){
+            if(nums[i] != nums[cur]){
+             cur++;
+             nums[cur] = nums[i];
+            }
+        }
+        return cur + 1;
+    }
+}
+/*思路：
+1、暴力解法，用两个变量进行存储，a存储中间值，b存储当前位置，a初始值为下标为0的数组元素值，b为0，for循环遍历数组，当发现数组值与变量值不同时，将当前数组值替换给a，数组下标为b+1的元素替换为a，b自增1，直到循环结束，返回b。
+2、经过优化之后，发现可以不需要使用两个变量来中间存储，直接用下标进行交换即可。
+*/