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leetcode 40. Combination Sum II (枚举子集，和为定值)

·106 words·1 min
ACM Leetcode 枚举子集
Note: This article is available in Chinese only. 本文暂无英文版本。 View original
  * Total Accepted: **106670**
  * Total Submissions: **329718**
  * Difficulty: **Medium**
  * Contributor: **LeetCode**

Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.

Each number in C may only be used once in the combination.

Note:

  * All numbers (including target) will be positive integers.
  * The solution set must not contain duplicate combinations.

题意：若干正数，求所有和为target的组合。

思路：枚举子集，用和剪枝。

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leetcode 90. Subsets II (枚举子集)

·266 words·1 min
ACM 枚举子集 Leetcode
Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,2], a solution is: 1[ 2 [2], 3 [1], 4 [1,2,2], 5 [2,2], 6 [1,2], 7 [] 8] 思路： 复习（？）一下 枚举子集的三种写法 （还有种更飘逸的…先不写了orz 这道题我用位向量法A的。。

leetcode 495. Teemo Attacking

·251 words·1 min
ACM Leetcode
In LLP world, there is a hero called Teemo and his attacking can make his enemy Ashe be in poisoned condition. Now, given the Teemo’s attacking ascending time series towards Ashe and the poisoning time duration per Teemo’s attacking, you need to output the total time that Ashe is in poisoned condition. You may assume that Teemo attacks at the very beginning of a specific time point, and makes Ashe be in poisoned condition immediately. 题意：若干长度相同的区间，升序给出区间左端点，以及区间长度。问区间被覆盖的长度总和。

leetcode 442. Find All Duplicates in an Array（找出出现两次的元素）

·234 words·1 min
ACM Leetcode
Given an array of integers, 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once. Find all the elements that appear twice in this array. Could you do it without extra space and in O(n) runtime? 思路：还是一个映射，如果某个位置要映射的时候已经为负了，就说明之前映射过该位置，那么该位置对应的元素就是出现了两个的元素。

leetcode 48. Rotate Image (旋转方阵(in place))

·575 words·2 mins
ACM Leetcode
You are given an n x n 2D matrix representing an image. Rotate the image by 90 degrees (clockwise). Follow up: Could you do this in-place? 题意：给一个n*n的方阵，要求顺时针旋转90度。 思路：(x,y)->(y,n-1-x); 要求in-place的做法的话，其实是若干长度为4的环，保护一个节点，然后顺次做就好了。

leetcode 54. Spiral Matrix (矩阵蛇形取数)

·291 words·1 min
ACM Leetcode
Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. 思路：。。。再次让我回想起高一的暑假。。。。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月11日 星期二 19时42分05秒 4File Name :54.cpp 5************************************************ */ 6class Solution { 7public: 8 int n,m; //0右，1下，2左，3上 9 int cal( int &x,int &y,int dir,vector<vector<bool> > & vis) 10 { 11 if (dir==0) 12 { 13 if (y<=n-2&&!vis[x][y+1]) y++; 14 else 15 { 16 dir++; 17 x++; 18 } 19 return dir; 20 } 21 if (dir==1) 22 { 23 if (x<=m-2&&!vis[x+1][y]) x++; 24 else 25 { 26 dir++; 27 y--; 28 } 29 return dir; 30 } 31 if (dir==2) 32 { 33 if (y>=1&&!vis[x][y-1]) y--; 34 else 35 { 36 dir++; 37 x--; 38 } 39 return dir; 40 } 41 if (dir==3) 42 { 43 if (x>=1&&!vis[x-1][y]) x--; 44 else 45 { 46 dir = 0 ; 47 y++; 48 } 49 return dir; 50 } 51 } 52 vector<int> spiralOrder(vector<vector<int>>& matrix) { 53 vector<int>res; 54 m = matrix.size(); 55 if (m==0) return res; 56 n = matrix[0].size(); 57 if (n==0) return res; 58 vector<vector<bool> >vis(m,vector<bool>(n,false)); 59 int x,y,dir; 60 x = y = dir = 0 ; 61 for ( int i = 0 ; i < n*m ; i++) 62 { 63 printf("x:%d y:%d\n",x,y); 64 res.push_back(matrix[x][y]); 65 vis[x][y] = true; 66 dir = cal(x,y,dir,vis); 67 } 68 return res; 69 } 70};