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Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Find all possible combinations of k numbers that add up to a number n, given that only numbers from 1 to 9 can be used and each combination should be a unique set of numbers.

Given two integers n and k, return all possible combinations of k numbers out of 1 … n. 思路：就是枚举子集，根据集合的大小剪枝。。。最后只要集合大小为k的集合

The set [1,2,3,…,_n_] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3):

Given a collection of numbers that might contain duplicates, return all possible unique permutations.__ 思路：和leet code 46 类似，最后用set去个重即可。。

Given a collection of distinct numbers, return all possible permutations. 思路：调用n-1次 leetcode 31 解题报告 中提到的算法即可。。。

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order).

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Given an array of integers sorted in ascending order, 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).

Given a set of candidate numbers (C) (without duplicates) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times.

* 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.

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.

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.

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。

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。。。。

Given a collection of intervals, merge all overlapping intervals. For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18]. 思路：dp[i]表示能否到达位置i…无脑dp即可。。。

Given a collection of intervals, merge all overlapping intervals. For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18]. 思路：扫一遍即可。。 代码实现 1/* *********************************************** 2Author :111qqz 3Created Time :2017年04月11日 星期二 19时15分30秒 4File Name :56.cpp 5************************************************ */ 6/** 7 8 * Definition for an interval. 9 10 * struct Interval { 11 12 * int start; 13 14 * int end; 15 16 * Interval() : start(0), end(0) {} 17 18 * Interval(int s, int e) : start(s), end(e) {} 19 20 * }; 21 22 */ 23 24class Solution { 25 26public: 27 28 int n; 29 static bool cmp(Interval A,Interval B) 30 { 31 return A.start<B.start; 32 } 33 vector<Interval> merge(vector<Interval>& pi) { 34 vector<Interval>res; 35 n = pi.size(); 36 if (n==0) return res; 37 sort(pi.begin(),pi.end(),cmp); 38 int l = -1,r = -1; 39 for ( int i = 0 ; i < n ; i++) 40 { 41 if (l==-1&&r==-1) 42 { 43 l = pi[0].start; 44 r = pi[0].end; 45 continue; 46 } 47 if (pi[i].start<=r) 48 { 49 r = max(r,pi[i].end); 50 continue; 51 } 52 if (pi[i].start>r) 53 { 54 res.push_back(Interval(l,r)); 55 l = pi[i].start; 56 r = pi[i].end; 57 continue; 58 } 59 } 60 //最后一组不要忘记 61 res.push_back(Interval(l,r)); 62 int siz = res.size(); 63 for ( int i = 0 ; i < siz ;i++) printf("%d ",res[i].start,res[i].end); 64 65 66 return res; 67 68 } 69 70};

Given an integer n, generate a square matrix filled with elements from 1 to _n_2 in spiral order. 思路：仿佛回到高一的那个暑假。。。