LeetCode--4sum

题目描述

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)

分析：和上一题一样，把一个数固定，然后就可以转换为三个数相加问题，然后再把一个数固定，就可以转换成两个数相加问题

class Solution {
public:
   vector<vector<int> > fourSum(vector<int> &num, int target)
   {
    int len = num.size();
    vector<vector<int>> res;
    if(len<4) return res;
    sort(num.begin(),num.end());
    for(int i=0;i<len;i++)
    {
        int threeSum = target-num[i];
        for(int j=i+1;j<len;j++)
        {
            int twoSum = threeSum-num[j];
            int left=j+1,right=len-1;
            while(left<right)
            {
                int curSum = num[left]+num[right];
                if(curSum < twoSum)
                    left++;
                else if(curSum > twoSum)
                    right--;
                else
                {
                    vector<int> temp={num[i],num[j],num[left],num[right]};
                    //temp[0] = num[i],temp[1]=num[j],temp[2]=num[left],temp[3]=num[right];
                    res.push_back(temp);
                    while(left<right && num[left]== temp[2])
                        ++left;
                    while(left<right && num[right]==temp[3])
                        --right;
                }
            }
            while(j<len-1 && num[j] == num[j+1])
                ++j;
        }
        while(i<len-1 && num[i]==num[i+1])
              ++i;
    }
    return res; 
  }
};