LeetCode-18. 4SumLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/4sum/

题目描述

Given an array nums of n integers and an integer target, are there elements a, b, c, and din 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]
]

解题思路

【C++解法】 

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> res;
        if(nums.size()<4) return res;
        sort(nums.begin(),nums.end());
        if(nums[0]+nums[1]+nums[2]+nums[3]>target || 
           nums[nums.size()-4]+nums[nums.size()-3]+nums[nums.size()-2]+nums[nums.size()-1]<target) return res;
        for(int i=0;i<nums.size()-3;i++){
            auto target3 = target-nums[i];
            if(nums[i+1]+nums[i+2]+nums[i+3]>target3) break;
            if(nums[nums.size()-3]+nums[nums.size()-2]+nums[nums.size()-1] < target3) continue;
            for(int j = i+1; j<nums.size()-2;j++){            
                auto target2 = target3-nums[j];
                if(nums[j+1]+nums[j+2]>target2) break;  
                if(nums[nums.size()-2]+nums[nums.size()-1]<target2) continue;
                int front = j+1, back = nums.size()-1;
                while(front<back) {
                    auto sum2 = nums[front]+nums[back];
                    if(sum2<target2) front++;
                    else if(sum2>target2) back--;
                    else {
                        res.push_back({nums[i],nums[j],nums[front++],nums[back--]});
                        while(front<back && nums[front-1]==nums[front]) front++;
                        while(front<back && nums[back+1]==nums[back]) back--;
                    }
                }
                while(j<nums.size()-2 && nums[j]==nums[j+1]) j++;
            }
            while(i<nums.size()-3 && nums[i]==nums[i+1]) i++;
        }
        return res;
    }
};