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

思路：

很新并且测试数量级很小的题，应该是某个contest的第一题。规定中其实就是把当前i分为奇偶，然后构造出一个数组，来找出这个数组中最大的值。感觉没有什么特别要注意的点，就是给定n小于等于1的时候额外判断以下即可。因此循环要从i=2开始，其他就是每次记录下当前的最大值，最后返回。

代码：

class Solution {
public:
    int getMaximumGenerated(int n) {
        if(n<=1)
            return n;
        vector<int> res={0,1};
        int ans=1;
        for(int i=2;i<=n;i++)
        {
            if(i%2==0)
                res.push_back(res[i/2]);
            else
                res.push_back(res[(i-1)/2]+res[(i-1)/2+1]);
            ans=max(ans,res.back());
        }
        return ans;
    }
};