18.4Sum (Map)

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)
思路: 用求3Sum的方法,外加一次for循环,时间复杂度O(n3),结果是Time limit exceeded

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int size = nums.size();
if(size < 3) return result;

sort(nums.begin(), nums.end());
for(int i = 0; i < size-3; i++){
if(i > 0 && nums[i]==nums[i-1])  continue;
for(int j = i+1; j < size-2; j++){
if(j> i+1 && nums[j]==nums[j-1]) continue;
find(nums, j+1, size-1, -nums[i]-nums[j], i, j);
}
}
return result;
}
void find(vector<int>& nums, int start, int end, int target, int& index1, int& index2){
int sum;
while(start<end){
sum = nums[start]+nums[end];
if(sum == target){
item.clear();
item.push_back(nums[index1]);
item.push_back(nums[index2]);
item.push_back(nums[start]);
item.push_back(nums[end]);
result.push_back(item);
do{
start++;
}while(start!= end && nums[start] == nums[start-1]);
do{
end--;
}while(end!=start && nums[end] == nums[end+1]);
}
else if(sum>target){
do{
end--;
}while(end!=start && nums[end] == nums[end+1]);
}
else{
do{
start++;
}while(start!= end && nums[start] == nums[start-1]);
}
}
}

private:
vector<vector<int>> result;
vector<int> item;
};

用hash table。O(N^2)把所有pair存入hash表，pair中两个元素的和就是hash值。那么接下来求4sum就变成了在所有的pair value中求 2sum，这个就成了线性算法了。所以整体上这个算法是O(N^2)+O(n) = O(N^2)。

class Solution {
public:
vector<vector<int>> fourSum(vector<int>& nums, int target) {
int size = nums.size();
int a, b, c, d;
vector<vector<int>> result;
unordered_map<int,vector<pair<int,int> > > mp;
unordered_map<int,int> cnt; //各个数的数量
if(size < 3) return result;

sort(nums.begin(), nums.end());
for(int i = 0; i < size-1; i++){
if(i > 0 && nums[i]==nums[i-1])  continue;
for(int j = i+1; j < size; j++){
if(j> i+1 && nums[j]==nums[j-1]) continue;
mp[nums[i]+nums[j]].push_back(pair<int,int>{nums[i],nums[j]});
}
}

for(int i = 0; i < size; i++){
cnt[nums[i]]++;
}

for(unordered_map<int,vector<pair<int,int> > >::iterator it1=mp.begin();it1!=mp.end();it1++){//遍历map
unordered_map<int,vector<pair<int,int> > >::iterator it2=mp.find(target - it1->first); //查找map
if(it2==mp.end()) continue;// not found
if(it1->first > it2->first) continue; //already checked，去重

for(int i = 0; i < it1->second.size(); i++){//访问map元素
for(int j = 0; j < it2->second.size(); j++){
a = it1->second[i].first; //访问pair元素
b = it1->second[i].second;
c = it2->second[j].first;
d = it2->second[j].second;

if(max(a,b) > min(c,d)) continue; //四个数两两组合，有6种情况，这里只取两个最小的数在it1的情况，去重
cnt[a]--;
cnt[b]--;
cnt[c]--;
cnt[d]--;
if(cnt[a]<0||cnt[b]<0||cnt[c]<0||cnt[d]<0){
cnt[a]++;
cnt[b]++;
cnt[c]++;
cnt[d]++;
continue;
}

cnt[a]++;
cnt[b]++;
cnt[c]++;
cnt[d]++;
vector<int> tmp = {a,b,c,d};
sort(tmp.begin(),tmp.end());
result.push_back(tmp);
}
}
}
return result;
}
};