题目如下:
You are given an integer
n
. A 0-indexed integer arraynums
of lengthn + 1
is generated in the following way:
nums[0] = 0
nums[1] = 1
nums[2 * i] = nums[i]
when2 <= 2 * i <= n
nums[2 * i + 1] = nums[i] + nums[i + 1]
when2 <= 2 * i + 1 <= n
Return the maximum integer in the array
nums
.
Example 1:
Input: n = 7 Output: 3 Explanation: According to the given rules: nums[0] = 0 nums[1] = 1 nums[(1 * 2) = 2] = nums[1] = 1 nums[(1 * 2) + 1 = 3] = nums[1] + nums[2] = 1 + 1 = 2 nums[(2 * 2) = 4] = nums[2] = 1 nums[(2 * 2) + 1 = 5] = nums[2] + nums[3] = 1 + 2 = 3 nums[(3 * 2) = 6] = nums[3] = 2 nums[(3 * 2) + 1 = 7] = nums[3] + nums[4] = 2 + 1 = 3 Hence, nums = [0,1,1,2,1,3,2,3], and the maximum is max(0,1,1,2,1,3,2,3) = 3.Example 2:
Input: n = 2 Output: 1 Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.Example 3:
Input: n = 3 Output: 2 Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.Constraints:
0 <= n <= 100
解题思路:反正n最大才100,挨个算吧。
代码如下:
class Solution(object): def getMaximumGenerated(self, n): """ :type n: int :rtype: int """ if n == 0:return 0 val = [0,1] + [0] * (n-1) for i in range(2,n+1): if i % 2 == 0: val[i] = val[i/2] else: val[i] = val[i/2] + val[i/2+1] return max(val)