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;
}