[Leetcode1_Two Sum]

两数之和算法解析

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本文介绍了一种经典的算法问题——寻找整型数组中和为目标数的两个元素的下标。提供了两种解决方案，一种是使用暴力循环实现，时间复杂度为O(n^2)；另一种是通过构建字典的方法来提高查找效率，将时间复杂度降低到接近O(n)。代码示例包括Python和C++两种语言。

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].

求整型数组中，和为某个数（target）的两个数的下标

解法一：暴力循环, 复杂度为O(n^2)
解法二：建立字典，利用字典的内置的查找函数降低复杂度

#PYTHON 代码:
class Solution():
    def TwoSum(nums, target):
        d = {}
        for i in range(len(nums)):
            if d[nums[i]] not in d:
                d[target-nums[i]] = i
            else:
                return d[nums[i]], i

//c++解法
#include <unordered_map>
class Solution{
    vector<int> TwoSum(vector<int> &nums, int target){
        std::unordered_map<int, int> d;
        vector<int> res;
        for(int i = 0; i < nums.size(); ++i){
            if(d.find(nums[i] == d.end()){
                d[target - nums[i]] = i;
            }else{
                res.push_back(d[nums[i]]);
                res.push_back(i);
                return res;
            }
        }
    }
}