LeetCode 1646. Get the maximum value in the generated array

1646. Get the maximum value in the generated array

Give you an integer n . According to the following rules to generate a length of n + 1 Array of nums ：

• nums[0] = 0
• nums[1] = 1
• When 2 <= 2 * i <= n when ,nums[2 * i] = nums[i]
• When 2 <= 2 * i + 1 <= n when ,nums[2 * i + 1] = nums[i] + nums[i + 1]

Returns the generated array nums Medium Maximum value .

Example 1：

 Input ：n = 7
 Output ：3
 explain ： According to 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
 therefore ,nums = [0,1,1,2,1,3,2,3], Maximum  3

Example 2：

 Input ：n = 2
 Output ：1
 explain ： According to rules ,nums[0]、nums[1]  and  nums[2]  The maximum of these is  1

Example 3：

 Input ：n = 3
 Output ：2
 explain ： According to rules ,nums[0]、nums[1]、nums[2]  and  nums[3]  The maximum of these is  2

Tips ：

• 0 <= n <= 100

Two 、 Method 1

simulation

class Solution {
    
    public int getMaximumGenerated(int n) {
    
        if (n == 0) {
    
            return 0;
        }
        int[] res = new int[n + 1];
        res[1] = 1;
        for (int i = 0; i < n; i++) {
    
            if (2 * i <= n) res[2 * i] = res[i];
            if (2 * i + 1 <= n) res[2 * i + 1] = res[i] + res[i + 1];
        }
        return Arrays.stream(res).max().getAsInt();
    }
}

Complexity analysis

• Time complexity ：O(n).

• Spatial complexity ：O(n).