Recca Chao 的 gitHub page

推廣網站開發,包含 Laravel 和 Kotlin 後端撰寫、自動化測試、讀書心得等。Taiwan Kotlin User Group 管理員。

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Hi, here's your problem today. This problem was recently asked by Facebook:

Starting at index 0, for an element n at index i, you are allowed to jump at most n indexes ahead. Given a list of numbers, find the minimum number of jumps to reach the end of the list.

Example: Input: [3, 2, 5, 1, 1, 9, 3, 4] Output: 2 Explanation:

The minimum number of jumps to get to the end of the list is 2: 3 -> 5 -> 4

Here's a starting point:

def jumpToEnd(nums):

Fill this in.

print jumpToEnd([3, 2, 5, 1, 1, 9, 3, 4])

2