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：

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
        vector<int> res;
        for (int i = 0; i < nums.size(); ++i) {
            m[nums[i]] = i;
        }
        for (int i = 0; i < nums.size(); ++i) {
            int t = target - nums[i];
            if (m.count(t) && m[t] != i) {
                res.push_back(i);
                res.push_back(m[t]);
                break;
            }
        }
        return res;
    }
};

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

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> m;
        for(int i = 0; i < nums.size(); ++ i)
        {
            int n = target - nums[i];
            if(m.count(n))
                return {i, m[n]};
            m[nums[i]] = i;
        }
        return {};
    }
};