Jump Game II Java

最新推荐文章于 2025-04-21 22:09:55 发布

原创最新推荐文章于 2025-04-21 22:09:55 发布 · 427 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #dp

LeetCode DP 专栏收录该内容

14 篇文章

订阅专栏

本文介绍了一个算法问题：如何在给定的非负整数数组中从第一个位置到达最后一个位置所需的最少跳跃次数。通过实例说明了算法的具体实现过程，并提供了一段Java代码作为解决方案。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.)

Extension of Jump Game

Optimize the number of step as well

now we have step & step -1

when step-1 can not reach current position i, then

we need step++, then make current position reachable.

public class Solution {
    public int jump(int[] A) {
        int len=A.length;
        if(len==0) return 0;
        int reach=0;
        int step=0;
        int lastReach=0;
        for(int i=0;i<=reach && i<len;i++) {
            //can not reach current position
            if (i > lastReach) {
                step++;
                lastReach = reach;
            }
            reach = Math.max(reach, A[i] + i);
        }
        return reach>=len-1?step:0;
    }
}