原题

Given an array nums of n integers and an integer target, are there elements abc, 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

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