LeetCode | 1. Two Sum

最新推荐文章于 2025-09-11 19:19:55 发布

原创最新推荐文章于 2025-09-11 19:19:55 发布 · 214 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #practice #编程练习

坚持就是胜利专栏收录该内容

10 篇文章

订阅专栏

本文深入探讨了经典的两数之和算法问题，提供了三种解决方案：暴力解题法、使用哈希表的两遍法和单遍哈希表法。通过对比不同方法的时间复杂度，帮助读者理解算法效率的重要性。

Description

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].

For Loops | 暴力解题法

用for loop遍历数组两次，O( $n^2$ )

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    	vector<int> result;
    	for (size_t i = 0; i < nums.size()-1; i++) {
    		int numToFind = target - nums[i];
    		
    		for (size_t j = i+1; j < nums.size(); j++) {
    			if (nums[j] == numToFind) {
    				result.push_back(i);
    				result.push_back(j);
    				break;
    			}
    		}
    		
    	}
    	return result;
    }
}

Hash Table | 聪明的哈希表

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    	Map<int,int> num_map;
    	vector<int> result;
    	
    	for (size_t i = 0; i < nums.size(); i++) {
    		//initialize the map with <number, index>
    		num_map[nums[i]] = i;
    	}
    	
    	for (size_t i = 0; i < nums.size(); i++) {
    		int numToFind = target - nums[i];
    		
    		if (num_map.find(numToFind) != num_map.end()) {
    			//found
    			if (i != num_map[numToFind]) {
    				//not the same index
    				result.push_back(i);
    				result.push_back(num_map[numToFind]);
    				break;
    			}
    		}
    	}
    	
    	return result;
    }
}

One Loop Hash Table | 精简哈希表

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
    	Map<int,int> num_map;
    	vector<int> result;
    	
    	for (size_t i = 0; i < nums.size(); i++) {
    		int numToFind = target - nums[i];
    		if (num_map.find(numToFind) != num_map.end()) {
    			//number found
    			result.push_back(i);
    			result.push_back(num_map[numToFind]);
    			break;
    		}
    		
    		//number not found, add to hash table
    		num_map[nums[i]] = i;
    	}
    	return result;
    }
}