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].
Boring brushing leetcode.
Found that the subject is a company interview topic.
This problem mainly use unordered_map .
unordered_map allow two or more apper in data.but is not ordered.
this is my solutation:
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> mp; vector<int> v; for (int i = 0; i < nums.size(); ++i ){ int x = target - nums[i]; if (mp.count(x) > 0) { v.push_back((mp.find(x))->second); v.push_back(i); break; } mp.insert({nums[i],i}); } return v; } };
o(n)
STL中,map对应的数据结构是红黑树。红黑树是一种近似于平衡的二叉查找树,里面的数据是有序的。在红黑树上做查找操作的时间复杂度为 O(logN)。而unordered_map对应 哈希表,哈希表的特点就是查找效率高,时间复杂度为常数级别 O(1), 而额外空间复杂度则要高出许多。所以对于需要高效率查询的情况,使用unordered_map容器。而如果对内存大小比较敏感或者数据存储要求有序的话,则可以用map容器。
原文地址:http://www.cnblogs.com/pk28/
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