[Leetcode]Jump Game II

最新推荐文章于 2021-06-29 09:54:08 发布

benjamin_sunny_li

最新推荐文章于 2021-06-29 09:54:08 发布

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本文链接：https://blog.youkuaiyun.com/weixin_38111819/article/details/80055543

本文介绍了一个巧妙的贪心算法，该算法能在O(N)的时间复杂度内找出从数组起始位置到末尾所需的最少跳跃次数。通过记录每一步能够达到的最远距离，算法有效地减少了不必要的跳跃。

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.

Example:

Input: [2,3,1,1,4]
Output: 2
Explanation: 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.

一种巧妙并且不太容易理解的贪心算法可以达到O(N)的时间复杂度，只不过在算法中要记录当前一跳所能到达的最远距离、上一跳所能到达的最远距离，和当前所使用跳数就可以了。

class Solution(object):
    def jump(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        ret = 0 #当前跳数
        last = 0 #上一跳可达最远距离
        cur = 0 #当前跳可达最远距离
        for i in range(len(nums)):
            if i > last: #需要进行下一次跳跃，更新last和跳数ret
                last = cur
                ret += 1
            cur = max(cur, i+nums[i]) #记录当前可达的最远点
        return ret