LeetCode OJ

题目：

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)

解题思路：

　　先对num排序，然后从两头枚举两个数，O(n^2)，后两个数在前两个数之间的两端开始，和小了左边的往右，和大了右边的往左调整，O(n)，总共O(n^3)。

代码：

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        sort(num.begin(), num.end());
        vector<vector<int>> ans;
        
        for (int left = 0; left < num.size(); ) {
            for (int right = num.size()-1; right > left+2; ) {
                int sum_of_two = num[left] + num[right];
                
                int ml = left + 1, mr = right - 1;
                while (ml < mr) {
                    if (sum_of_two + num[ml] + num[mr] == target) {
                        vector<int> cur_ans = {num[left], num[ml], num[mr], num[right]};
                        ans.push_back(cur_ans);
                        ml++;
                        mr--;
                    } else if (sum_of_two + num[ml] + num[mr] < target) {
                        ml ++;
                    } else {
                        mr --;
                    }
                    //去重
                    while (ml < mr && (ml != left + 1 && num[ml] == num[ml-1])) ml ++;
                    while (ml < mr && (mr != right - 1 && num[mr] == num[mr+1])) mr--;
                }
                for(right --; right > left + 2 && num[right] == num[right + 1]; right --);
            }
            for(left ++; left < num.size() - 3 && num[left] == num[left - 1]; left ++);
        }
        return ans;
    }
};