标签：提交 else imu 数组 ini max tree ret lex
最近做的题记录下。
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
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 };
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].
[5, 7, 7, 8, 8, 10] and target value 8,[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
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 }
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.
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。
Add Digits, Maximum Depth of BinaryTree, Search for a Range, Single Number,Find the Difference
标签：提交 else imu 数组 ini max tree ret lex
原文地址：http://www.cnblogs.com/tenxi/p/6034976.html
