[Leetcode1_Two Sum]
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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; } } }}
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