57 leetcode - Jump Game II

最新推荐文章于 2025-08-05 16:42:55 发布

忧桑的小兔子

最新推荐文章于 2025-08-05 16:42:55 发布

阅读量263

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/lis_12/article/details/53871880

算法专栏收录该内容

84 篇文章

订阅专栏

本文介绍了一种解决Jump Game II问题的算法实现，旨在帮助读者理解如何在给定的非负整数数组中，从第一个索引位置出发，利用每个元素代表的最大跳跃长度，以最少的跳跃次数达到数组的最后一个索引。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#!/usr/bin/python
# -*- coding: utf-8 -*-
'''
Jump Game II
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.
假设每一个都能到达最后一个索引
'''
class Solution(object):
    def jump(self, nums):
        """
        超时了...醉了...
        :type nums: List[int]
        :rtype: int
        """
        length = len(nums)
        if length < 2:
            return 0

        min_step = [length] * length
        min_step[0],min_step[1] = 0,1
        for index,val in enumerate(nums):
            for jump in range(index + 1,index + val + 1):
                tmp = min_step[index] + 1
                if jump < length and min_step[jump] > tmp:
                    min_step[jump] = tmp

        return min_step[-1]

    def jump(self, nums):
        """
        type nums: List[int]
        rtype: int
        """
        length = len(nums)
        if length < 2:
            return 0

        if (nums[0] + 1) >= length:
            return 1

        max_dis1 = max_dis2 = nums[0]
        target = length - 1
        min_step = 1
        index = 1
        while index < length:    #寻找每一步能走的最大值
            if index <= max_dis1:#在走一步的范围内,查看下一步最远走到哪里
                max_dis2 = max(max_dis2,index + nums[index])
                if (max_dis2 + 1) >= length:
                    return min_step + 1
                index += 1
            else:                #超过了这步的最远距离,步数加一,更新成下一步能走的最远距离
                min_step += 1
                max_dis1 = max_dis2

if __name__ == "__main__":
    s = Solution()
    print s.jump([10,4,2])
    print s.jump([2,3,0,1,4])
    print s.jump([1,1,1,1,1])
    print s.jump([1,1,2,1,1])