leetcode/1646.Get-Maximum-in-Generated-Array/README.md
You are given an integer n. An array nums of length n + 1 is generated in the following way:
nums[0] = 0nums[1] = 1nums[2 * i] = nums[i] when 2 <= 2 * i <= nnums[2 * i + 1] = nums[i] + nums[i + 1] when 2 <= 2 * i + 1 <= nReturn ****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 :
返回生成数组 nums 中的 最大值。
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
}