leetcode-1646-easy

Get Maximum in Generated Array

You are given an integer n. A 0-indexed integer 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 max(0,1,1,2,1,3,2,3) = 3.
Example 2:

Input: n = 2
Output: 1
Explanation: According to the given rules, nums = [0,1,1]. The maximum is max(0,1,1) = 1.
Example 3:

Input: n = 3
Output: 2
Explanation: According to the given rules, nums = [0,1,1,2]. The maximum is max(0,1,1,2) = 2.
Constraints:

0 <= n <= 100

思路一：遍历

    public int getMaximumGenerated(int n) {
        if (n < 2) {
            return n;
        }
        
        int[] nums = new int[n + 1];

        nums[0] = 0;
        nums[1] = 1;
        for (int i = 2; i <= n; i++) {
            if ((i & 1) == 1) {
                int x = (i - 1) / 2;
                nums[i] = nums[x] + nums[x + 1];
            } else {
                nums[i] = nums[i / 2];
            }
        }

        int max = 0;
        for (int num : nums) {
            max = Math.max(max, num);
        }

        return max;
    }