18. 4Sum

Posted 2020-10-14 hozhangel

---

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: 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]
]
找出四个数，相加等于零。

在一个列表里寻找四个数的情况，遍历方法一般为：

for（int i = 0； i < totalNum - 3； i++）{
    for(int m = i + 1; m < totalNum - 2 ; m++){
    　　int j = m+1;
    　　int k = totalNum - 1;

        while(j<k){

    　　//接着看情况使  j++ 或  k--
        }
    //不想重复组合时：这里要判断num[m] 是否等于 num[m+1],等于的话m++,跳过下一个元素
    }
    //同理，不想重复组合时：这里要判断num[i] 是否等于 num[i+1],等于的话i++,跳过下一个元素
}

和3sum closet  、 3sum 方法类似

 

最终提交正确的代码：

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
    
        vector<vector<int> > res;
        if(nums.size() < 4) return res;
        std::sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 3; i++) {
            for(int m = i + 1; m < nums.size() - 2; m ++){
                
                int target1 = target -nums[i] - nums[m];
                int front = m + 1;
                int back = nums.size() - 1;

                while (front < back) {

                    int sum = nums[front] + nums[back];

                    if (sum < target1)
                        front++;

                    else if (sum > target1)
                        back--;
                    else {
                        vector<int> triplet(4, 0);
                        triplet[0] = nums[i];
                        triplet[1] = nums[m];
                        triplet[2] = nums[front];
                        triplet[3] = nums[back];
                        res.push_back(triplet);

                        while (front < back && nums[front] == triplet[2]) front++;

                        while (front < back && nums[back] == triplet[3]) back--;
                    }

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