Leecode #18 4Sum

一、 问题描述

Leecode第十八题，题目为：
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]
]
问题理解为：

给定一个包含n个整数的数组和一个整数目标，nums中是否有a、b、c和d元素使得a + b + c + d = target?找出数组中四数之和为给定目标和的所有唯一组。
注意:

解集不能重复。

例:

给定数组nums = [1,0， - 1,0， - 2,2]， target = 0。
解为:
[
[- 1, 0, 0, 1]，
[-2， -1, 1, 2]，
[- 2, 0, 0, 2]
]

二、算法思路
1、
2、

三、实现代码

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int> > res;
        if(nums.empty()) return res;
        sort(nums.begin(), nums.end());
        
        for(int i = 0; i < nums.size(); i++)
        {
        	int target_three = target - nums[i];
        	for(int j = i + 1; j < nums.size(); j++)
        	{
        		 int target_two = target_three - nums[j];
				 int front = j + 1;
				 int back = nums.size() - 1;
				 
				 while(front < back)
				 {
				 	int two_sum = nums[front] + nums[back];
				 	if(two_sum < target_two) front++;
				 	else if(two_sum > target_two) back--;
				 	else
				 	{
				 		vector<int> quadruplet(4, 0);
                        quadruplet[0] = nums[i];
                        quadruplet[1] = nums[j];
                        quadruplet[2] = nums[front];
                        quadruplet[3] = nums[back];
                        res.push_back(quadruplet);
                                           
                        while(front < back && nums[front] == quadruplet[2]) front++;
					}
				 } 

				 while(j + 1 < nums.size() && nums[j + 1] == nums[j]) j++;
			}
			
			
			while(i + 1 < nums.size() && nums[i + 1] == nums[i]) i++;
		}
		
		return res;
    }
};