LeetCode18. 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: 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，定义4个指针，p1,p2,p3,p4.顺序遍历p1，p2=p1+1开始遍历，然后p3=p2+1,p4=n-1，sum4 =nums[p1]+nums[p2]+nums[p3]+nums[p4]，当 sum4小于target时，p3右移，sum4大于target时，p4左移，sum4等于target时，加入vector,直到p3>=p4

#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int n=nums.size();
vector<vector<int>> res;
if(n<4) return res;
vector<int> resb;
sort(nums.begin(),nums.end());
int e1=n-3,e2=n-2,p1=0,p2=1,p3=1,p4=n-1,sum4=0;
while(p1<e1){
p2 = p1+1;
while(p2<e2){
p3 = p2+1;
p4 = n-1;
while(p3<p4){
sum4 =nums[p1]+nums[p2]+nums[p3]+nums[p4];
if(sum4<target){
++p3;
}
else if(sum4>target){
--p4;
}
else{
resb.push_back(nums[p1]);
resb.push_back(nums[p2]);
resb.push_back(nums[p3]);
resb.push_back(nums[p4]);
res.push_back(resb);
resb.clear();
if(nums[p3]==nums[p3+1]){
while(nums[p3]==nums[p3+1]){
++p3;
}
++p3;
}
else{
++p3;
}
--p4;
}
}
if(nums[p2]==nums[p2+1]){
while(nums[p2]==nums[p2+1]){
++p2;
}
++p2;
}
else{
++p2;
}
}
if(nums[p1]==nums[p1+1]){
while(nums[p1]==nums[p1+1]){
++p1;
}
}
++p1;
}
return res;
}
};
void main(){
int a[8]={-3,-2,-1,0,0,1,2,3};
vector<int> nums;
//nums.reserve(8);
//nums.assign(&a[0],&a[8]);
nums.push_back(-3);
nums.push_back(-2);
nums.push_back(-1);
nums.push_back(0);
nums.push_back(0);
nums.push_back(1);
nums.push_back(2);
nums.push_back(3);

vector<int>::iterator it3=nums.begin();
for(it3=nums.begin();it3!=nums.end();++it3){
cout<<(*it3)<<" ";
}
cout<<endl;

Solution So;

vector<vector<int>> res=So.fourSum(nums,0);
vector<vector<int>>::iterator it1=res.begin();
vector<int>::iterator it2=(*it1).begin();
for(it1=res.begin();it1!=res.end();++it1){
for(it2=(*it1).begin();it2!=(*it1).end();++it2){
cout<<(*it2)<<" ";
}
cout<<endl;
}
}