[Leetcode] Two Sum
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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].
分析
给定一个整型数组,找到两个数,使得它们的和为 target,返回这两个数的索引(假设有且仅有一组解)。
很自然的一个想法是暴力搜索[代码1],但是时间复杂度是
另外一种思路[代码2]:使用 map,建立起 “数值-索引”的对应关系。首先遍历一遍数组,完成 map 的初始化,之后再遍历一遍数组,同时在 map 中查找 target-nums[i],找到则返回对应索引,时间复杂度
代码1
class Solution {public: vector<int> twoSum(vector<int>& nums, int target) { for (int i = 0; i < nums.size(); i++) { for (int j = i + 1; j < nums.size(); j++) { if (nums[i] + nums[j] == target) return{i, j}; } } return {}; }};
代码2
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 {}; }};
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