01Leetcode-TwoSum

题目描述：

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 = [2, 7, 11, 15], target = 9, 
Because nums[0] + nums[1] = 2 + 7 = 9, 
return [0, 1]. 

题意：给定目标值，在数组中查找到两数满足两数相加等于目标值，返回两数索引,此题已经说明每个目标值都只有唯一一组数符合 
时间复杂度：                   暴力搜索：O（n^2） 
                             采用哈希搜索法：O（n） 
python可利用字典,建立的索引为key-value:数值-索引 

python版---map

class Solution(object):
    def  twoSum(self, nums, target):
        dict = {}
        for i in range(len(nums)):
            if target - nums[i] in dict   #  target-nums[i]为key ( 索引 下标 )
                return [dict[target - num[i]], i]
            dict[nums[i]] = i
        return  []

# 字典中的key < == > 列表、元组、字符串 中的[下标、索引]

C++版---map

用map实现，建立元素与下标的映射！ 找到和为target的元素，返回下标

vector <int> twoSum(vector<int> nums, int target)
{
    vector<int> res;
    map<int, int> m;
    for(int i=0; i< nums.size(); i++)
    {
        if(m.find(target - num[i]!=m.end())
            {
                res.push_back(m[target-nums[i]]);
                res.push_back(i);
            }
            m[nums[i]] = i;
    }
    return res;
}

reference ：

python: https://blog.csdn.net/qq_28119401/article/details/52972461