内容简介：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。时间复杂度：O(n^3)空间复杂度：O(n)
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。
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)
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