最近做的题记录下。
258. Add Digits
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
1 int addDigits(int num) {
2 char strint[12] = {0};
3 sprintf(strint, "%d", num);
4 int i=0,a = 0;
5 while(strint[i+1]!= ‘\0‘)
6 {
7 a = (strint[i]-‘0‘)+(strint[i+1]-‘0‘);
8 if(a>9)
9 {
10 a = a%10+a/10;
11 }
12 strint[++i] = a+‘0‘;
13 }
14 return strint[i]-‘0‘;
15 }
上边这是O(n)的复杂度。网上还有O(1)的复杂度的：return (num - 1) % 9 + 1;
104. Maximum Depth of Binary Tree
求二叉树的深度，可以用递归也可以用循环，如下：
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * struct TreeNode *left;
6 * struct TreeNode *right;
7 * };
8 */
9 int maxDepth(struct TreeNode* root) {
10 int left = 0,right = 0;
11 if (root == NULL) return 0;
12
13 left = maxDepth(root->left);
14 right = maxDepth(root->right);
15
16 return left>right? left+1:right+1;
17 }
******************************************************************************************************************
用队列存结点，记录每一层的结点数levelCount，每出一个结点levelCount--，如果减为0说明要进入下一层，然后depth++，同时队列此时的结点数就是下一层的结点数。
1 /**
2 * Definition for a binary tree node.
3 * struct TreeNode {
4 * int val;
5 * TreeNode *left;
6 * TreeNode *right;
7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
8 * };
9 */
10 class Solution {
11 public:
12 int maxDepth(TreeNode* root) {
13 if (root == NULL) return 0;
14
15 //深度和每一层节点数
16 int depth = 0,levelCount = 1;
17 //队列保存每一层的节点数
18 queue<TreeNode*> node;
19
20 node.push(root);
21 while(!node.empty())
22 {
23 //依次遍历每一个结点
24 TreeNode *p = node.front();
25 node.pop();
26 levelCount--;
27
28 if(p->left) node.push(p->left);
29 if(p->right) node.push(p->right);
30
31 if (levelCount == 0)
32 {
33 //保存下一层节点数，深度加1
34 depth++;
35 levelCount = node.size();
36 }
37 }
38 return depth;
39 }
40 };
34. Search for a Range
Given a sorted array of integers, find the starting and ending position of a given target value.
Your algorithm‘s runtime complexity must be in the order of O(log n).
If the target is not found in the array, return [-1, -1].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4]
1 vector<int> searchRange(vector<int>& nums, int target) {
2 int size = nums.size();
3 int l=0,r=size-1;
4 while(l<=r)
5 {
6 int mid = (l+r)/2;
7 if (nums[mid] >= target)
8 {
9 r = mid-1;
10 }
11 else if(nums[mid] < target)
12 {
13 l = mid+1;
14 }
15 }
16 int left = l;
17 l = 0;
18 r = size-1;
19 while(l<=r)
20 {
21 int mid = (l+r)/2;
22 if (nums[mid] <= target)
23 {
24 l = mid+1;
25 }
26 else if(nums[mid] > target)
27 {
28 r = mid-1;
29 }
30 }
31 int right = r;
32 vector<int> v;
33 v.push_back(left);
34 v.push_back(right);
35 if(nums[left] != target || nums[right] != target)
36 {
37 v[0] = -1;
38 v[1] = -1;
39 }
40 return v;
41 }
这个题就是要把握好边界，比如找左边界时 =号要给右边界的判断，找右边界则相反。这样才能保证不遗漏
136. Single Number
Given an array of integers, every element appears twice except for one. Find that single one.
直接用异或
1 int singleNumber(vector<int>& nums) {
2 int result=0;
3 int len = nums.size();
4 if (len <=0) return 0;
5 for(int i=0;i<len;i++)
6 {
7 result ^= nums[i];
8 }
9 return result;
10 }
389. Find the Difference
Given two strings s and t which consist of only lowercase letters.
String t is generated by random shuffling string s and then add one more letter at a random position.
Find the letter that was added in t
这种方法挺慢，我提交显示16ms，没有那种用char数组来模拟hash快。
1 char findTheDifference(string s, string t) {
2 map<char, int> sumChar;
3 char res;
4 for(auto ch: s) sumChar[ch]++;
5 for(auto ch: t)
6 {
7 if(--sumChar[ch]<0)
8 res = ch;
9 }
10 return res;
11 }
这道题也可以用异或 因为相同的会异或掉，剩下一个就是多余的char。