[leetcode55]跳跃游戏(超简单)

最新推荐文章于 2024-08-07 21:49:38 发布

SmallNancy

最新推荐文章于 2024-08-07 21:49:38 发布

阅读量353

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签： Java leetcode

本文链接：https://blog.youkuaiyun.com/SmallNancy/article/details/100071184

leetcode 专栏收录该内容

1 篇文章

订阅专栏

本文深入解析了一种称为跳跃游戏的算法问题，探讨了如何判断能否从数组的起始位置到达最后位置。通过示例说明了算法的实现过程，包括如何计算最大跳跃长度和更新可达范围。

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.
Determine if you are able to reach the last index.
Example 1:
Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
jump length is 0, which makes it impossible to reach the last index.

package com.leetcode;

/**
 * @author King
 * @date 2019/8/26 8:56
 */
public class 跳跃游戏 {
    public boolean canJump(int[] nums) {
        int digist = 0; // 下标值
        for (int i = 0; i < nums.length && i <= digist ;i++) {
            digist = Math.max(digist,nums[i]+i);
        }
        return digist >= nums.length - 1;

    }
}