LeetCode -- Two Sum

最新推荐文章于 2025-08-23 17:53:17 发布

转载最新推荐文章于 2025-08-23 17:53:17 发布 · 50 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/little-YTMM/p/4786233.html

文章标签：

#runtime

本文解析了给定整数数组中寻找两个数使它们加和等于特定目标数的问题。介绍了两种解决方案：一种是时间复杂度为O(n²)的暴力求解方法；另一种是时间复杂度为O(n)，使用哈希表降低查找复杂度的方法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Question:

Given an array of integers, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2

Analysis:

O(n2) runtime, O(1) space – Brute force:

The brute force approach is simple. Loop through each element x and find if there is another value that equals to target – x. As finding another value requires looping through the rest of array, its runtime complexity is O(n2).

用暴力求解方法，时间复杂度较高，两次for循环，但方法容易想到。

O(n) runtime, O(n) space – Hash table:

We could reduce the runtime complexity of looking up a value to O(1) using a hash map that maps a value to its index.

用HashMap记录遍历过的数字所需要的val，然后返回index即可，时间复杂度小。

Answer：

1. 暴力求解：

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] res = new int[2];
        
        for(int i=0; i<nums.length; i++) {
            for(int j=i+1; j<nums.length; j++) {
                int sum = nums[i] + nums[j];
                if(sum == target) {
                    res[0] = i + 1;
                    res[1] = j + 1;
                    break;
                }
            }
        }
        return res;
    }
}

2. HashMap巧解。

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        
        int[] res = new int[2];
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for(int i=0; i<nums.length; i++) {
            if(map.containsKey(nums[i])) {
                res[0] = map.get(nums[i]) + 1;
                res[1] = i + 1;
            }
            else {
                map.put(target - nums[i], i);
            }
        }
        
        
        return res;
    }
}