每日算法之十六：4sum

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)

直接上代码：

最主要的还是进行去重。

<span style="font-size:18px;">class Solution {  public:    vector<vector<int> > fourSum(vector<int> &num, int target) {         vector<vector<int> > ret;        if(num.size()==0) return ret;        sort(num.begin(),num.end());        for(int i = 0;i<num.size();++i)        {            if(i>0&&num[i] == num[i-1])                continue;            for(int j = i+1;j<num.size();j++)            {                if(j>i+1&&num[j] == num[j-1])                    continue;                int k = j+1,t = num.size()-1;                while(k<t)                {                    if(k>j+1&&num[k] == num[k-1])//这里是if，而不是while.如果是while可能会越界，在每一次                        {k++;continue;}     //指针变化之后都要进行判断。避免越界的发生。</span><pre name="code" class="cpp"><span style="font-size:18px;">               //这里是contimue，而不是break.如果是while,也不能是break，因为可能导致t指针的重复。</span>

if(t<num.size()-1&&num[t] == num[t+1]) {t--;continue;} // int sum = num[i]+num[j]+num[k]+num[t]; if(sum<target) k++; else if(sum>target) t--; else { vector<int> v; v.push_back(num[i]); v.push_back(num[j]); v.push_back(num[k]); v.push_back(num[t]); ret.push_back(v); k++;//哪一个变化都可以 } } } } return ret; }};