1. 题目
Given an array S of n integers, are there elements a, b, c, and d in S 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.
For example, given array S = [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]
]
2. 思路
和3 Sum closeset一样的,只是多遍历了一层。
3. 代码
耗时:123ms
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> ret;
int size = nums.size();
switch(size) {
case 0:
case 1:
case 2:
case 3:
return ret;
}
sort(nums.begin(), nums.end());
for (int i = 0; i < size - 3; i++) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < size - 2; j++) {
if (j > i + 1 && nums[j] == nums[j-1]) continue;
int k = j + 1;
int l = size - 1;
while (k < l) {
int sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum == target) {
bool same = false;
if (ret.size() > 0) {
vector<int>& last = ret[ret.size() - 1];
if (last[0] == nums[i] && last[1] == nums[j] && last[2] == nums[k] && last[3] == nums[l]) {
same = true;
}
}
if (!same) {
vector<int> got;
got.push_back(nums[i]);
got.push_back(nums[j]);
got.push_back(nums[k]);
got.push_back(nums[l]);
ret.push_back(got);
}
k++;
while(k < l && nums[k] == nums[k-1])k++;
l--;
while (k < l && nums[l] == nums[l+1])l--;
} else if (sum < target) {
k++;
while(k < l && nums[k] == nums[k-1])k++;
} else {
l--;
while (k < l && nums[l] == nums[l+1])l--;
}
}
}
}
return ret;
}
};
**粗体** _斜体_ [链接](http://example.com) `代码` - 列表 > 引用
。你还可以使用@
来通知其他用户。