[LeetCode] 1646. Get Maximum in Generated Array

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

获取生成数组中的最大值。

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。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/get-maximum-in-generated-array
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这道题不涉及算法，就是根据题意把nums数组模拟出来，然后在过程中一直记录最大值是什么即可。

时间O(n)

空间O(n)

Java实现

class Solution {
    public int getMaximumGenerated(int n) {
        // corner case
        if (n <= 1) {
            return n;
        }
        
        // normal case
        int[] nums = new int[n + 1];
        int max = 0;
        nums[0] = 0;
        nums[1] = 1;
        for (int i = 2; i < nums.length; i++) {
            nums[i] = nums[i / 2];
            if (i % 2 == 1) {
                nums[i] += nums[i / 2 + 1];
            }
            max = Math.max(max, nums[i]);
        }
        return max;
    }
}

LeetCode 题目总结