4.30更新,已经AC
18. 4Sum My Submissions QuestionEditorial Solution
Total Accepted: 71102 Total Submissions: 299393 Difficulty: Medium
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:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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)
结果：
282 / 282 test cases passed.
Status: Accepted
Runtime: 76 ms
beats:64.95%
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int> > result;
sort(nums.begin(),nums.end());
int n=nums.size();
for(int i=0;i<n-1;++i){
for(int j=i+1;j<n;++j){
int front = nums[i]+nums[j];
int beg = j+1,end = n-1;
while(beg<end){
int tsum = front + nums[beg]+nums[end];
if(tsum==target){
vector<int> vec;
vec.push_back(nums[i]),vec.push_back(nums[j]);
vec.push_back(nums[beg]),vec.push_back(nums[end]);
result.push_back(vec);
while(++beg<end&&nums[beg-1]==nums[beg]);
while(--end>beg&&nums[end+1]==nums[end]);
}
else {
if(tsum<target)beg++;
else end--;
}
}
}
}
sort(result.begin(),result.end());
result.erase(unique(result.begin(),result.end()),result.end());
return result;
}
};思路：先保存两个整数的和，然后两层循环搞定
平均时间复杂度：
还是time limit,期待更好的解法。。。。
class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
vector<vector<int> > result;
if(nums.size()<4)return result;
sort(nums.begin(),nums.end());
map<int,vector<pair<int,int> > > cache;
for(size_t a=0;a<nums.size();++a){
for(size_t b=0;b<nums.size();++b){
cache[nums[a]+nums[b]].push_back(pair<int,int>(a,b));
}
}
for(int c=0;c<nums.size();++c){
for(size_t d=c+1;d<nums.size();++d){
const int key = target -nums[c]-nums[d];
if(cache.find(key)==cache.end())continue;
const auto&vec =cache[key];
for(size_t k=0;k<vec.size();++k){
if(c<=vec[k].second)continue;
result.push_back({
nums[vec[k].first],nums[vec[k].second],nums[c],nums[d]});
}
}
}
sort(result.begin(),result.end());
result.erase(unique(result.begin(),result.end()),result.end());
return result;
}
};testcase: