JAVA代码—算法基础:跳跃游戏(II)

本文介绍了一种解决跳跃游戏II问题的算法实现。该问题要求找到从数组起始位置到达末尾所需的最少跳跃次数,每个元素代表当前位置的最大跳跃长度。通过设定当前区域范围并逐步更新边界来寻找最优解。

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

跳跃游戏(Jump Game II)

问题描述:

Given an array of non-negative integers, you are initially positioned at the first index of the array.
Each element in the array represents your maximum jump length at that position.
Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Note:
You can assume that you can always reach the last index.

算法设计:

package com.bean.algorithmbasic;

public class JumpGameII {

    public static int jump(int[] nums) {
        // If nums.length < 2, means that we do not
        // need to move at all.
        if (nums == null || nums.length < 2) {
            return 0;
        }

        // First set up current region, which is
        // from 0 to nums[0].
        int l = 0;
        int r = nums[0];
        // Since the length of nums is greater than
        // 1, we need at least 1 step.
        int step = 1;

        // We go through all elements in the region.
        while (l <= r) {

            // If the right of current region is greater
            // than nums.length - 1, that means we are done.
            if (r >= nums.length - 1) {
                return step;
            }

            // We should know how far can we reach in current
            // region.
            int max = Integer.MIN_VALUE;
            for (; l <= r; l++) {
                max = Math.max(max, l + nums[l]);
            }

            // If we can reach far more in this round, we update
            // the boundary of current region, and also add a step.
            if (max > r) {
                l = r;
                r = max;
                step++;
            }
        }

        // We can not finish the job.
        return -1;
    }

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int [] arrayDemo= {2,3,1,1,4};
        int result=jump(arrayDemo);
        System.out.println("ANSWER is: "+result);
    }

}

(完)

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