标签：des style blog io ar color os sp for
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:
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)
C++代码实现：
#include<iostream> #include<algorithm> #include<vector> using namespace std; class Solution { public: vector<vector<int> > fourSum(vector<int> &num,int target) { if(num.empty()) return vector<vector<int> >(); sort(num.begin(),num.end()); vector<vector<int> > ret; int n=num.size(); int i,j; for(i=0; i<n-3; i++) { //只保留第一个不重复的，其余的都删了,因为left会选择重复的 if(i>=1&&num[i]==num[i-1]) continue; for(j=n-1; j>i+2; j--) { //只保留最后一个不重复的，其余的都删了,因为right会选择重复的 if(j<n-1&&num[j+1]==num[j]) continue; int left=i+1; int right=j-1; vector<int> tmp; while(left<right) { //只保留最后一个不重复的，其余的都删了,因为left会选择重复的 if(right<j-1&&num[right]==num[right+1]) right--; else if(num[i]+num[j]+num[left]+num[right]==target) { tmp= {num[i],num[left],num[right],num[j]}; ret.push_back(tmp); left++; right--; } else if(num[i]+num[j]+num[left]+num[right]<target) left++; else if(num[i]+num[j]+num[left]+num[right]>target) right--; } } } return ret; } }; int main() { vector<int> vec= {-5,5,4,-3,0,0,4,-2}; Solution s; vector<vector<int> > result=s.fourSum(vec,4); for(auto a:result) { for(auto v:a) cout<<v<<" "; cout<<endl; } cout<<endl; }
注意处理重复的vector。。。。
标签：des style blog io ar color os sp for
原文地址：http://www.cnblogs.com/wuchanming/p/4120872.html
