https://leetcode.com/problemset/algorithms/上面的题目，每天做几道题目，大体从准确率高至低做下去
编程语言为c语言，因为跑的最快…
| 237 | Delete Node in a Linked List | 47.8% | Easy |
Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node
with value 3, the linked list should become 1
-> 2 -> 4 after calling your function.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node) {
if(node->next != NULL){
node->val = node->next->val;
node->next=node->next->next;
}
}删除给定的节点，只需要将当前节点的val改成下个节点的值，这时候相当于有两个相同值的节点，再将当前节点的next指针指向下下个节点，也就是跳过第二个相同的节点就好了。
| 104 | Maximum Depth of Binary Tree | 45.1% | Easy |
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
int maxDepth(struct TreeNode* root) {
if(root == NULL) // 递归出口
return 0;
int depthLeft = maxDepth(root->left);
int depthRight = maxDepth(root->right);
return depthLeft > depthRight ? (depthLeft + 1) : (depthRight + 1);
}求二叉树的深度，使用递归的思想，将root想做普通节点，任何节点的深度都是左右节点较深的那个值+1，除了树叶（深度为0，也是递归的结束条件）
| 136 | Single Number | 45.0% | Medium |
Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
int singleNumber(int* nums, int numsSize) {
if (nums == 0 || numsSize < 1)
return 0;
int key = nums[0];
for (int i = 1; i < numsSize; ++i) {
key ^= nums[i];
}
return key;
}难点在于数组规模增大，运行时间只能线性增长。
将数组的所有元素进行异或运算，相同的元素异或等于0，唯一的元素与0异或等于自己。
| 100 | Same Tree | 41.5% | Easy |
Given two binary trees, write a function to check if they are equal or not.
Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* struct TreeNode *left;
* struct TreeNode *right;
* };
*/
bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
if(!p && !q) return true; //NULL together, the same,skip the remaining code
if(!p || !q) return false; //one is NULL but the other is not
return (p->val == q->val) && isSameTree(p->left, q->left) && isSameTree(p->right, q->right);
}跟104题类似，使用递归的思想，二叉树相同，意味着从根节点开始，必须符合对应节点的值相同+左节点为根节点的树相同+右节点为根节点的树相同
先判断!p&&!q同时为树叶节点，两个为相同的树
如果不同时为树叶节点，则判断两者是否有一方是树叶节点，如果是，则他们不是一样的树
对根节点运行算法就可以得出结果
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怒刷leetcode的题目（1）237、104、136、100
原文地址：http://blog.csdn.net/edwardwayne/article/details/46971767
