title
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]
]
solution
if (j == i+1 || nums[j] != nums[j - 1]) 使得后面的相等的都是紧挨的
给一个包含n个数的整数数组S，在S中找到所有使得和为给定整数target的四元组(a, b, c, d)。
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ans;
sort(nums.begin(), nums.end());
for (int i = 0; i < nums.size(); i++) {
if (i == 0 || nums[i] != nums[i - 1]) {
for (int j = i + 1; j < nums.size(); j++) {
if (j == i+1 || nums[j] != nums[j - 1]) {
int sum1 = nums[i] + nums[j];
int l = j + 1, r = nums.size() - 1;
while (l < r) {
int sum = sum1 + nums[l] + nums[r];
if (sum == target) {
ans.push_back({ nums[i],nums[j],nums[l],nums[r] });
while (l < r&&nums[l] == nums[l + 1]) l++;
while (l < r&&nums[r] == nums[r - 1]) r--;
l++;
r--;
}
else if (sum < target) {
l++;
}
else {
r--;
}
}
}
}
}
}
return ans;
}
};