Merge pull request #398 from plin005/master

039 to commit week002 homework (LeetCode_242 & LeetCode_671)

Author: GeekUniversity

Week_01/id_39/LeetCode_083_039.java

@@ -0,0 +1,28 @@
+package com.paula.algorithmsAndDataStructure.leetCode_83;
+
+public class LeetCode_083_039 {
+	public ListNode deleteDuplicates(ListNode head) {
+		if(head == null) return null;
+		
+		ListNode p = head;
+		ListNode q = null;
+        while(p.next != null) {
+        	if(p.val == p.next.val) {
+        		if(q == null || (q != null && q.val != p.val)) {
+        			q = p;
+        		}
+        	}else {
+        		if(q != null && q.val == p.val) {
+        			q.next = p.next;
+        		}
+        	}
+        	p = p.next;
+        }
+        
+        if(q != null && p.val == q.val && p.next == null) {
+        	q.next = null;
+        } 
+		return head;
+    }
+
+}

Week_01/id_39/LeetCode_922_039.java

@@ -0,0 +1,42 @@
+package com.paula.algorithmsAndDataStructure;
+
+public class LeetCode_922_039 {
+	public int[] sortArrayByParityII(int[] A) {
+        int len = A.length;
+    	int i=0; //偶数下标
+    	int j=i+1; //奇数下标
+    	Boolean evenIndexOddValue = false;
+    	Boolean oddIndexEvenValue = false;
+    	
+        while (i < len-1 && j < len) {
+        	evenIndexOddValue = isOdd(A[i]);
+            oddIndexEvenValue= isEven(A[j]);
+
+        	if(!evenIndexOddValue) i +=2;
+        	if(!oddIndexEvenValue) j +=2;
+        	if(evenIndexOddValue && oddIndexEvenValue) {
+        		int tempVal = A[i];
+        		A[i] = A[j];
+        		A[j] = tempVal;
+        		
+        		i += 2;
+        		j += 2;
+        		
+        	}
+        }
+        
+     
+    	return A;
+    }
+    
+    public  boolean isEven(int value) {
+    	
+    	return value%2 == 0 ? true : false;
+    }
+    
+    public  boolean isOdd(int value) {
+    	
+    	return value%2 == 1 ? true : false;
+    }
+
+}

Week_02/id_39/LeetCode_242_039.java

@@ -0,0 +1,49 @@
+
+import java.util.HashMap;
+import java.util.Map;
+
+/**
+  * 
+  * @author Paula
+  *
+  * 
+  */
+ 
+class Solution {
+	private static void chechString(String s, Map<Character, Integer> smap) {
+        for (int i = 0; i < s.length(); i++) {
+            if (smap.containsKey(s.charAt(i))) {
+                smap.put(s.charAt(i), smap.get(s.charAt(i)) + 1);
+            } else {
+                smap.put(s.charAt(i), 1);
+            }
+        }
+    }
+
+	public static boolean isAnagram(String s, String t) {
+        if (s.equals(t)) {
+            return true;
+        } else if (null == s || null == t || s.length() != t.length() || s.length() <= 0) {//如果长度不相等，直接返回
+            return false;
+        }
+        Map<Character, Integer> smap = new HashMap<Character, Integer>();
+        chechString(s, smap);
+        for (int i = 0; i < t.length(); i++) {
+        	char tmp = t.charAt(i);
+            if (smap.containsKey(tmp)) {
+                smap.put(tmp, smap.get(tmp) - 1);
+            }
+            if (0 == smap.get(tmp)) {
+                smap.remove(tmp);
+            }
+        }
+        
+        return smap.size() == 0;
+    }
+
+	public static void main(String[] args) {
+		//("anagram", new HashMap<Character, Integer>());
+		System.out.println(isAnagram("anagram", "nagaram"));
+		System.out.println(isAnagram("rat", "car"));
+	}
+}

Week_02/id_39/LeetCode_671_039.java

@@ -0,0 +1,51 @@
+
+ /**
+  * 
+  * @author Paula
+  *
+  * 执行结果:
+  * 执行用时 : 1 ms, 在Second Minimum Node In a Binary Tree的Java提交中击败了86.02% 的用户
+  * 内存消耗 : 34 MB, 在Second Minimum Node In a Binary Tree的Java提交中击败了77.61% 的用户
+  * 
+  * test case:
+  * [2,2,5,null,null,5,7]
+  * [1,1,3,1,1,3,4,3,1]
+  * [5,8,5]
+  */
+ 
+class Solution {
+    public int findSecondMinimumValue(TreeNode root) {
+    	if(root == null) return -1;
+    	//第二小节点值
+    	int secondMin = -1;
+    	
+    	TreeNode p = root; //第二小的节点
+    	
+    	//检查左节点
+    	if (p.left != null) {
+    		//如果左子节点值不等于当前节点值, 则左子节点值必然大于当前节点的值
+    		if(p.left.val != p.val) secondMin = p.left.val;
+    		//如果左子节点值不等于当前节点值, 继续遍历左子节点下的分支
+        	else secondMin = findSecondMinimumValue(p.left);
+    	}
+    	
+    	//检查右节点
+    	if (p.right != null) {
+    		//如果右子节点值不等于当前节点值, 则右子节点值必然大于当前节点的值
+    		if(p.right.val != p.val) {
+    			//对比从左子树得到的secondMin和右子节点的值
+        		if(secondMin == -1 || (secondMin != -1 && p.right.val < secondMin) )
+        			secondMin = p.right.val;
+        		
+            //如果右子节点值不等于当前节点值, 继续遍历右子节点下的分支
+    		} else {
+    			int tmp = findSecondMinimumValue(p.right);
+    			if (tmp !=-1 && secondMin !=-1 && tmp < secondMin) 
+    				secondMin = tmp;
+    		}
+    		
+    	}
+    	   
+    	return secondMin;
+    }
+}