題目:
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:
3-sum基礎上外加一層循環
代碼:
class Solution {
public:
vector<vector<int> > fourSum(vector<int>& nums, int target) {
vector<vector<int> > vec;
int Len = nums.size();
if(Len == 0)return vec;
sort(nums.begin(), nums.end());
vector<int> v(4);
for(int i = 0; i < Len-3; ++i){
if(i > 0 && nums[i-1] == nums[i])continue;
if(nums[i] + nums[i+1] + nums[i+2] + nums[i+3] > target)return vec;
else if(nums[i] + nums[Len-3] + nums[Len-2] + nums[Len-1] < target)continue;
v[0] = nums[i];
for(int j = i+1; j < Len-2; ++j){
if(j > i+1 && nums[j-1] == nums[j])continue;
if(nums[i] + nums[j] + nums[j+1] + nums[j+2] > target)break;
else if(nums[i] + nums[j] + nums[Len-2] + nums[Len-1] < target)continue;
v[1] = nums[j];
for(int L = j+1, R=Len-1; L < R;){
if(L > j+1 && nums[L-1] == nums[L]){
++L;
continue;
}
if(nums[i] + nums[j] + nums[L]+nums[R] == target){
v[2] = nums[L];
v[3] = nums[R];
vec.push_back(v);
if(nums[L] != nums[R]){
while(L < R && nums[++L] == nums[L-1]){};
while(L < R && nums[--R] == nums[R+1]){};
}else{
break;
}
}else if(nums[i] + nums[j] + nums[L]+nums[R] > target){
--R;
}else{
++L;
}
}
}
}
return vec;
}
};