Leetcode题解-18. 4Sum

Leetcode题解-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]

]

思路

把4Sum问题化成3Sum，3Sum化成2Sum，所以还是在2Sum的基础上加两重循环。2Sum的思路是：把数组nums从小到大排序，一个指针front指向数组头，一个指针back指向数组尾，如果front和back指向数组元素之和小于target，front++，如果大于target则back–，直到front不小于back结束

代码

vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int>> res;
int l = nums.size();
if(l < 4) return res;
sort(nums.begin(), nums.end());
for(int i = 0; i < l-3; i++){
int ttarget = target - nums[i];
for(int j = i+1; j < l-2; j++){
int front = j + 1, back = l - 1, tttarget = ttarget - nums[j];
while(front < back){
int sum = nums[front] + nums[back];
if(sum < tttarget) front++;
else if(sum > tttarget) back--;
else{
vector<int> tem(4,0);
tem[0] = nums[i];
tem[1] = nums[j];
tem[2] = nums[front];
tem[3] = nums[back];
res.push_back(tem);
while(front < back && nums[front] == tem[2]) front++;
while(front < back && nums[back] == tem[3]) back--;
}
}
while(nums[j] == nums[j+1]) j++;
}
while(nums[i] == nums[i+1]) i++;
}
return res;
}