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].

C++:

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

python:
1.两次遍历，第一次遍历找出数组下标和值的对应关系，存到dict里面。第二次遍历nums[i]，找target-nums[i]是否在dict。

enumerate() 函数用于将一个可遍历的数据对象(如列表、元组或字符串)组合为一个索引序列，同时列出数据和数据下标，一般用在 for 循环当中。

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        res_dict={}
        if len(nums)<2:
            pass
        for i,num in enumerate(nums):
            res_dict[num]=i
        for i, num in enumerate(nums):
            if target - num in res_dict and res_dict[target - num] != i:
                return [i, res_dict[target - num]]

2.一次遍历：

class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        res_dict={}
        if len(nums)<2:
            pass
        for i,num in enumerate(nums):
            if target-num in res_dict:
                return [res_dict[target-num],i]
            else:
                res_dict[num]=i
        return [0,0]