LeetCode Everyday: 1. Two Sum

本文介绍了一种高效解决两数之和问题的算法。通过使用哈希表记录已访问过的数字及其索引,该算法能在O(n)的时间复杂度内找到数组中两个数相加等于特定目标值的元素索引。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >


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].

UPDATE (2016/2/13):

The return format had been changed to zero-based indices. Please read the above updated description carefully.

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


Using a HashMap to keep tracking what number has been visited, O(n) solution

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值