1646. Get Maximum in Generated Array

# 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 #

• 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]

## 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

func getMaximumGenerated(n int) int {
if n == 0 {
return 0
}
nums, max := make([]int, n+1), 0
nums[0], nums[1] = 0, 1
for i := 0; i <= n; i++ {
if nums[i] > max {
max = nums[i]
}
if 2*i >= 2 && 2*i <= n {
nums[2*i] = nums[i]
}
if 2*i+1 >= 2 && 2*i+1 <= n {
nums[2*i+1] = nums[i] + nums[i+1]
}
}
return max
}
``````
Nov 24, 2020