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1646.Get-Maximum-in-Generated-Array

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1646. Get Maximum in Generated Array

题目

You are given an integer n. An array nums of length n + 1 is generated in the following way:

  • nums[0] = 0
  • nums[1] = 1
  • nums[2 * i] = nums[i] when 2 <= 2 * i <= n
  • nums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 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 3.

Example 2:

Input: n = 2 Output: 1 Explanation: According to the given rules, the maximum between nums[0], nums[1], and nums[2] is 1.

Example 3:

Input: n = 3 Output: 2 Explanation: According to the given rules, the maximum between nums[0], nums[1], nums[2], and nums[3] is 2.

Constraints:

  • 0 <= n <= 100

题目大意

给你一个整数 n 。按下述规则生成一个长度为 n + 1 的数组 nums ：

  • nums[0] = 0
  • nums[1] = 1
  • 当 2 <= 2 * i <= n 时，nums[2 * i] = nums[i]
  • 当 2 <= 2 * i + 1 <= n 时，nums[2 * i + 1] = nums[i] + nums[i + 1]

返回生成数组 nums 中的 最大值。

解题思路

  • 给出一个 n + 1 的数组，并按照生成规则生成这个数组，求出这个数组中的最大值。
  • 简单题，按照题意生成数组，边生成边记录和更新最大值即可。
  • 注意边界条件，当 n 为 0 的时候，数组里面只有一个元素 0 。

代码

package leetcode funcgetMaximumGenerated(nint) int { ifn==0 { return0 } nums, max:=make([]int, n+1), 0nums[0], nums[1] =0, 1fori:=0; i<=n; i++ { ifnums[i] >max { max=nums[i] } if2*i>=2&&2*i<=n { nums[2*i] =nums[i] } if2*i+1>=2&&2*i+1<=n { nums[2*i+1] =nums[i] +nums[i+1] } } returnmax }
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