leetcode 18. 4Sum 双指针

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.

Example:

Given array nums = [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]
]

这个和 3Sum 的做法类似。在每层循环的开始可以进行一个判断，假如在此情况下的最小和都大于 target 则 break，若最大和都小于 target 则 continue

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        int n=nums.size();
        if(n<4)return res;
        sort(nums.begin(), nums.end());
        for(int i=0;i<n-3;i++){
            if(i>0&&nums[i]==nums[i-1])continue;
            int cur=nums[i];
            if(cur+nums[i+1]+nums[i+2]+nums[i+3]>target)break;
            if(cur+nums[n-1]+nums[n-2]+nums[n-3]<target)continue;
            for(int j=i+1;j<n-2;j++){
                if(j>i+1&&nums[j]==nums[j-1])continue;
                cur=nums[i]+nums[j];
                if(cur+nums[j+1]+nums[j+2]>target)break;
                if(cur+nums[n-1]+nums[n-2]<target)continue;
                int l=j+1,r=n-1;
                while(l<r){
                    cur=nums[i]+nums[j]+nums[l];
                    if(cur+nums[l+1]>target)break;
                    if(cur+nums[n-1]<target){
                        while(l<r&&nums[l]==nums[l+1])l++;l++;
                        continue;
                    }
                    cur+=nums[r];
                    if(cur==target){
                        res.push_back(vector<int>{nums[i],nums[j],nums[l],nums[r]});
                        while(l<r&&nums[l]==nums[l+1])l++;l++;
                    }else if(cur>target){while(l<r&&nums[r]==nums[r-1])r--;r--;}
                    else {while(l<r&&nums[l]==nums[l+1])l++;l++;}
                }
            }
        }
        return res;
    }
};