LeetCode 1. Two Sum

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, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

先遍历一遍数组，建立hash表，再遍历一遍数组，找出下标

c++代码：

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         unordered_map<int, int> m;
 5         vector<int> res;
 6         for (int i = 0; i < nums.size(); ++i) {
 7             m[nums[i]] = i;
 8         }
 9         for (int i = 0; i < nums.size(); ++i) {
10             int t = target - nums[i];
11             if (m.count(t) && m[t] != i) {
12                 res.push_back(i);
13                 res.push_back(m[t]);
14                 break;
15             }
16         }
17         return res;
18     }
19 };

其实第一遍遍历数组的时候就可以判断了，这样可以优化代码为

c++代码

 1 class Solution {
 2 public:
 3     vector<int> twoSum(vector<int>& nums, int target) {
 4         unordered_map<int, int> m;
 5         for(int i = 0; i < nums.size(); ++ i)
 6         {
 7             int n = target - nums[i];
 8             if(m.count(n))
 9                 return {i, m[n]};
10             m[nums[i]] = i;
11         }
12         return {};
13     }
14 };