[LeetCode]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)

4sum,可以归纳为3sum问题，这样复杂度 O(N^3)。但是实际中可以通过一些细节处理提高遍历无用解的情形。
16ms

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ret;
int len = nums.size();
sort(nums.begin(),nums.end());
for(int i=0;i<len-3;++i){
if(i>0&&nums[i]==nums[i-1]) continue;//dup start
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;//no result
if(nums[i]+nums[len-1]+nums[len-2]+nums[len-3]<target) continue;//at this start no result,go to next one,important
for(int j=i+1;j<len-2;++j){//3 sum
if(j>i+1&&nums[j]==nums[j-1]) continue;//dup start
if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break;//优化的关键
if(nums[i]+nums[j]+nums[len-1]+nums[len-2]<target) continue; //优化的关键
int left = j+1;
int right = len-1;
while(left<right){
if(nums[i]+nums[j]+nums[left]+nums[right]==target){
ret.push_back({nums[i],nums[j],nums[left],nums[right]});
++left;
--right;
while(nums[left]==nums[left-1]&&nums[right]==nums[right+1]){ //重复子解,重要
++left;
--right;
}
}
else if(nums[i]+nums[j]+nums[left]+nums[right]<target){
++left;
while(nums[left]==nums[left-1]){//重复解，重要
++left;
}
}
else if(nums[i]+nums[j]+nums[left]+nums[right]>target){
--right;
while(nums[right]==nums[right+1]){ //重复解
--right;
}
}
}
}
}
return ret;
}
};