原题
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] ]
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。
[code lang=”cpp”]
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;
}
};
[/code]
时间复杂度：O(n^3)
空间复杂度：O(n)
原题：https://leetcode.com/problems/4sum/description/
文章来源：胡小旭 => [LeetCode]4Sum