leetcode之4Sum

紧接上篇的3Sun问题，4Sum问题的描述如下：

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.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {1 0 -1 0 -2 2}, and target = 0.    A solution set is:    (-1,  0, 0, 1)    (-2, -1, 1, 2)    (-2,  0, 0, 2)

思路在上一篇中已经说的比较明确了，这篇就直接上代码吧。

class Solution {public:    vector<vector<int>>vv;    vector<vector<int> > fourSum(vector<int> &num, int target) {        if(num.size() < 4)return vv;vector<int>v;sort(num.begin(),num.end());for(int i = 0; i < num.size() - 3; i++){v.push_back(num[i]);ThreeSum(num,v,i + 1,target - num[i]);v.pop_back();}return vv;    }void ThreeSum(vector<int>num,vector<int>v,int index,int target){for(int i = index; i < num.size() - 2; i++){v.push_back(num[i]);twoSum(num,v, i+ 1,num.size() - 1,target - num[i]);v.pop_back();}}void twoSum(vector<int>num,vector<int>v,int left,int right,int target){while(left < right){int sum = num[left] + num[right];if(sum == target){v.push_back(num[left]);v.push_back(num[right]);if(find(vv.begin(),vv.end(),v) == vv.end()){vv.push_back(v);}v.pop_back();v.pop_back();left++;right--;}else if(sum < target)left++;elseright--;}}};

通过这道题，我更加深刻的认识到编码来不得半点儿马虎，任何的细节忽视都会导致致命的错误，在这道题中应该注意数组的进入和退出、遍历时的下标以及传参等问题。