Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [, , , ], target = ,

Because nums[] + nums[] =  +  = ,
return [, ].

1/ 首先想到的是 通过两次循环去寻找 这两个数,找到后立刻返回。

但是当提交运行的时候,会报错,运行时间过长。

2/ 想到的另一种方法是,先通过排序(nlgn),然后通过两个指针去前后遍历数组(n)

3/ 最后一种方法在网上看到的,因为自己对hashmap并不是很熟悉。一下贴出网上的hashmap的代码

class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
vector<int> res;
int length = numbers.size();
map<int,int> mp;
int find;
for(int i = ; i < length; ++i){
// if have target-numbers[i] return target-numbers[i] ,else create a target-numbers[i] element
find=mp[target - numbers[i]];
if( find ){
res.push_back(find);
res.push_back(i+);
break;
}
mp[numbers[i]] = i;
}
return res;
}
};

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