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Description:
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] ]
思路:这类题统称K-Sum,思路都特别相似;
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int> > res;
int n = nums.size();
if (n < 4)
return res;
sort(nums.begin(), nums.end());
for(int i = 0; i < n-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[n-3] + nums[n-2] + nums[n-1] < target)
continue;
for(int j = i+1; j < n-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[n-2] + nums[n-1] < target)
continue;
int l = j+1, r = n-1;
while(l < r) {
int tempSum = nums[i] + nums[j] + nums[l] + nums[r];
if(tempSum < target)
l ++;
else if(tempSum > target)
r --;
else {
res.push_back(vector<int> {nums[i], nums[j], nums[l], nums[r]});
do {
l ++;
} while(nums[l] == nums[l-1] && l < r);
do {
r --;
} while(nums[r] == nums[r+1] && l < r);
}
}
}
}
return res;
}
};