//解法一,map将数组分成n,0,p三组,寻找9(0,0,0),(n,0,p)(n,n,p)和(p,p,n)的组合,但是因为不可重复,所以最后两种组合比较麻烦,复杂度是O(n^2),遍历多次,最后两个case TLE
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
map<int,int> ne;
map<int,int> po;
int nne=0,npo=0,zero=0;
vector<vector<int>> answer;
for(int i=0;i<nums.size();i++){
if(nums[i]>0){
po[nums[i]]++;
npo++;
}
else if(nums[i]<0){
ne[nums[i]]++;
nne++;
}
else zero++;
}
if(zero>=3){
vector<int> v;
v.push_back(0);
v.push_back(0);
v.push_back(0);
answer.push_back(v);
}
if(nne&&npo){
if(zero){
for(map<int,int>::iterator it=ne.begin();it!=ne.end();it++){
if(po[-it->first]){
vector<int> v;
v.push_back(it->first);
v.push_back(0);
v.push_back(-it->first);
answer.push_back(v);
}
else po.erase(-it->first);
}
}
if(nne>=2){
for(map<int,int>::iterator i=ne.begin();i!=ne.end();i++){
for(map<int,int>::iterator j=i;j!=ne.end();j++){
if(i->first==j->first && i->second<2) continue;
else{
int left= -((i->first)+(j->first));
if(po[left]){
vector<int> v;
v.push_back(i->first);
v.push_back(j->first);
v.push_back(left);
answer.push_back(v);
}
else po.erase(left);
}
}
}
}
if(npo>=2){
for(map<int,int>::iterator i=po.begin();i!=po.end();i++){
for(map<int,int>::iterator j=i;j!=po.end();j++){
if(i->first==j->first && i->second<2) continue;
else{
int left=-(i->first+j->first);
if(ne[left]){
vector<int> v;
v.push_back(i->first);
v.push_back(j->first);
v.push_back(left);
answer.push_back(v);
}
else ne.erase(left);
}
}
}
}
}
return answer;
}
};
//解法2,每有一个数,去找符合条件的另外两个数。把数组排好序,可以利用二分法的思想,避免枚举所有其他两个数的组合。排序O(nlogn),后面的遍历约为O(n^2),
但是常数系数为1,而且没有map的cost
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
sort(nums.begin(),nums.end());
vector<vector<int>> result;
for(int i=0;i<nums.size();i++){
if(nums[i]>0) break;
if(i>0 && nums[i]==nums[i-1]) continue;
int left=i+1,right=nums.size()-1;
while(left<right){
int sum=nums[left]+nums[right]+nums[i];
if(sum>0){
right--;
}
else if(sum<0){
left++;
}
else{
result.push_back({nums[i],nums[left],nums[right]});
int l=nums[left];
int r=nums[right];
while(left<right&&nums[left]==l) left++;
while(right>left&&nums[right]==r) right--;
}
}
}
return result;
}
};