[LeetCode]4Sum

原题

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]
]

题解

本题的思路和3Sum一致，无非是多加一层循环。当然，时间复杂度就为O(n^3)了。

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector< vector<int> > rs;
        sort(nums.begin(), nums.end());
        int l = nums.size(), f = 0, b = 0;
        for (int i = 0; i < l - 3; ++i) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1; j < l - 2; j++) {
                f = j + 1;
                b = l - 1;
                if (j - i > 1 && nums[j] == nums[j - 1]) continue;
                while (f < b) {
                    if (nums[i] + nums[j] + nums[f] + nums[b] < target) {
                        while (nums[f] == nums[++f]) {}
                    } else if(nums[i] + nums[j] + nums[f] + nums[b] > target) {
                        while (nums[b] == nums[--b]) {}
                    } else {
                        vector<int> tmp = {nums[i], nums[j], nums[f], nums[b]};
                        rs.push_back(tmp);
                        while (nums[f] == nums[++f]) {}
                        while (nums[b] == nums[--b]) {}
                    }
                }
            }
        }
        return rs;
    }
};

时间复杂度：O(n^3)

空间复杂度：O(n)