Leetcode-Two Sum

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

思路：固定一位，找剩下的元素，判断其和是否等于target

时间复杂度O(n^2)

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int> res;
for(int i=0;i<numbers.size();i++)
for(int j=i+1;j<numbers.size();j++)
{
if(numbers[i]+numbers[j]==target)
{
res.push_back(i+1);
res.push_back(j+1);
return res;
}
}
return res;
}
};

提交的结果为：Time Limit Exceeded

下面做些优化：

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int> res;
res.clear();
multimap<int,int> m;
multimap<int,int>::iterator mit;
multimap<int,int>::iterator mit2;
int i;
for(i=0;i<numbers.size();i++){
//m[numbers[i]]=i+1;
m.insert(pair<int,int>(numbers[i],i+1));

}
for(mit=m.begin();mit!=m.end();mit++)
{
if((mit2=m.find(target-mit->first))!=m.end()&&mit2!=mit)
{
if(mit->second>mit2->second)
{
res.push_back(mit2->second);
res.push_back(mit->second);
}
else
{
res.push_back(mit->second);
res.push_back(mit2->second);
}
return res;
}
}

return res;
}
};

find函数的底层实现是红黑树，时间复杂度为O(logn)

所以整个函数的时间复杂度为O(nlogn)