LeetCode中的bug！！！！！！

1. Two Sum

第36行本来应该是if (nodes.get(left).key < nodes.get(right).key) { 一不小心误写成了if (nodes.get(left).key < nodes.get(right).value) { 竟然A了！！！！

 1 class Node{
 2     int key;
 3     int value;
 4     public Node(int key, int value) {
 5         this.key = key;
 6         this.value = value;
 7     }
 8 }
 9 public class Solution {
10     /*
11      * @param numbers : An array of Integer
12      * @param target : target = numbers[index1] + numbers[index2]
13      * @return : [index1 + 1, index2 + 1] (index1 < index2)
14      */
15     public int[] twoSum(int[] numbers, int target) {
16         if (numbers == null || numbers.length <= 1) {
17             return new int [2];
18         }
19         //put the key and value in list, the type of list is Node
20         List<Node> nodes = new ArrayList<>();
21         for (int i = 0; i < numbers.length; i++) {
22             nodes.add(new Node(i, numbers[i]));
23         }
24         //sort the list
25         Collections.sort(nodes, new Comparator<Node>() {
26             public int compare(Node a, Node b) {
27                 return a.value - b.value;
28             }
29         });
30         // Two points
31         int left = 0;
32         int right = numbers.length - 1;
33         int[] result = new int[2];
34         while (left < right) {
35             if (nodes.get(left).value + nodes.get(right).value == target) {
36                 if (nodes.get(left).key < nodes.get(right).value) { //第二个明显应该是key,但是第一次不小心写成了value，居然A了，但是lintcode却不能A
37                     result[0] = nodes.get(left).key;
38                     result[1] = nodes.get(right).key;
39                 } else {
40                     result[0] = nodes.get(right).key;
41                     result[1] = nodes.get(left).key;
42                 }
43                 return result;
44             } else if (nodes.get(left).value + nodes.get(right).value < target) {
45                 left++;
46             } else {
47                 right--;
48             }
49         }
50         return result;
51     }
52 }

572. Subtree of Another Tree

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public boolean isSubtree(TreeNode root1, TreeNode root2) {
12         //如果root2不进行判断，也A了，但是很明显这样是错的
13         //这样的话并没有考虑，两个都为空，以及root1不为空，而root2为空的情况
14         if (root2 == null) {
15             return true;
16         }    
17         if (root1 == null) {
18             return false;
19         }
20         if(isEqual(root1, root2)) {
21             return true;
22         }
23         return isSubtree(root1.left, root2) || isSubtree(root1.right, root2);
24     }
25     public boolean isEqual(TreeNode root1, TreeNode root2) {
26         if (root1 == null || root2 == null) {
27             return root1 == root2;
28         }
29         if (root1.val != root2.val) {
30             return false;
31         }
32         return isEqual(root1.left, root2.left) && isEqual(root1.right, root2.right);
33     }
34 }