Leetcode_4sum

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地址：https://oj.leetcode.com/problems/4sum/

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：跟3sum的思路一样http://blog.csdn.net/flyupliu/article/details/24438255

先排个序

主要是先定首尾两数，中间俩数分别向中间搜索。

代码中四个数的下标分别是i , j , k, m

去重比较麻烦, j 和 k 分别如果存在相邻的相同数肯定可以continue，但是i和m不行，没有多想直接在ans里find了, sorry方法这么low

参考代码：

class Solution {public:vector<vector<int> > fourSum(vector<int> &num, int target) {vector<vector<int>>ans;if(num.size()<4)return ans;sort(num.begin(), num.end());vector<int>vec(4, 0);for(int i = 0; i<(int)num.size()-3; ++i){//if(i && num[i]==num[i-1])//continue;int m = i + 3;for(;m<num.size(); ++m){//if(m > i+3 && num[m]==num[m-1])//continue;int j = i + 1, k = m-1;for(; j < k;){if(j > i + 1 && num[j]==num[j-1] ){++j;continue;}if(k < m-1 && num[k]==num[k+1]){--k;continue;}if(num[i]+num[j]+num[k]+num[m]<target)++j;else if(num[i]+num[j]+num[k]+num[m]>target)--k;else{vec[0] = num[i];vec[1] = num[j];vec[2] = num[k];vec[3] = num[m];if(find(ans.begin(), ans.end(), vec)==ans.end())    ans.push_back(vec);--k;++j;}}}}return ans;}};

SECOND TRIAL

为毛比上一个方法要慢100+ms

class Solution {public:    vector<vector<int> > fourSum(vector<int> &num, int target) {        vector<vector<int>>ans;        if(num.size()<4)            return ans;        sort(num.begin(), num.end());        int left = 0, right = 0;        vector<int>vec(4, 0);        for(int i = 0; i < num.size()-3; ++i)        {            if(i && num[i]==num[i-1])                continue;            for(int j = num.size()-1; j > i+2; --j)            {                if(j < num.size()-1 && num[j]==num[j+1])                    continue;                left = i + 1;                right = j - 1;                while(left < right)                {                       if(left > i+1 && num[left]==num[left-1])                        ++left;                    else if(right < j-1 && num[right]==num[right+1])                        --right;                    else if(num[i]+num[j]+num[left]+num[right]==target)                    {                        vec[0] = num[i];                        vec[1] = num[left];                        vec[2] = num[right];                        vec[3] = num[j];                        ans.push_back(vec);                        ++left;                        --right;                    }                    else if(num[i]+num[left]+num[right]+num[j]<target)                        ++left;                    else if(num[i]+num[left]+num[right]+num[j]>target)                        --right;                }            }        }        return ans;    }};

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