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]
]
题目大意:给一个数组和target,求数组中所有使a+b+c+d = target的可能。
题目思路:和3sum思路类似,构建4个索引,时间复杂度为O(n³).
代码如下:
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int l,m,n,k,sum;
int len = nums.size();
vector<vector<int>> ans;
sort(nums.begin(), nums.end());
for(int l = 0; l < len-3; l++){
if(nums[l]+nums[l+1]+nums[l+2]+nums[l+3] > target)
break;
if(nums[len-4]+nums[len-3]+nums[len-2]+nums[len-1] < target)
break;
for(m = l+1; m < len-2; m++){
int start = m+1, end = len-1;
// cout<<"start = "<<start<<" "<<"end = "<<end<<endl;
while(start < end){
sum = nums[l] + nums[m] + nums[start] + nums[end];
if(sum < target){
++start;
}
if(sum > target){
--end;
}
if(sum == target){
// cout<<"start = "<<start<<endl;
// cout<<"end = "<<end<<endl;
ans.push_back({nums[l] , nums[m] , nums[start] , nums[end]});
++start;
while(nums[start] == nums[start-1] && start < end)
++start;
while(nums[end] == nums[end+1] && start < end)
--end;
}
}
while(m < len-2 && nums[m] == nums[m+1])
m++;
// cout<<"now m = "<<m<<endl;
/* */
}
while(l < len-3 && nums[l] == nums[l+1])
++l;
}
return ans;
}
};