leetcode 18. 4Sum

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]
]

解：这一题跟3sum差不多，就是在外面在遍历一层，但是要先排序，然后注意递增的判断条件

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
int len = nums.size();
if(len < 4) return res;
std::sort(nums.begin(), nums.end());
for(int i = 0; i < len - 3; ++i){
if(i > 0 && nums[i] == nums[i - 1]) continue;
if(nums[i] + nums[i + 1] + nums[i + 2] + nums[i + 3] > target) break;
if(nums[i] + nums[len - 1] + nums[len - 2] + nums[len - 3] < target) continue;
for(int j = i+1; j < len - 2; ++j){
if(j > i + 1 && nums[j] == nums[j - 1]) continue;
if(nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) break;
if(nums[i] + nums[j] + nums[len - 1] + nums[len - 2] < target) continue;
int front = j + 1, back = len - 1;
while(front < back){
int sum = nums[i] + nums[j] + nums[front] + nums[back];
if(sum < target) front++;
else if(sum > target) back--;
else{
res.push_back(vector<int>{nums[i],nums[j],nums[front],nums[back]});
front++; back--;
while(front < back && nums[front - 1] == nums[front]) front++;
while(front < back && nums[back + 1] == nums[back]) back--;
}
}
}
}
return res;
}
};