Your task is to form an integer array nums
from an initial array of zeros arr
that is the same size as nums
.
Return the minimum number of function calls to make nums
from arr
.
The answer is guaranteed to fit in a 32-bit signed integer.
Example 1:
Input: nums = [1,5] Output: 5 Explanation: Increment by 1 (second element): [0, 0] to get [0, 1] (1 operation). Double all the elements: [0, 1] -> [0, 2] -> [0, 4] (2 operations). Increment by 1 (both elements) [0, 4] -> [1, 4] -> [1, 5] (2 operations). Total of operations: 1 + 2 + 2 = 5.
Example 2:
Input: nums = [2,2] Output: 3 Explanation: Increment by 1 (both elements) [0, 0] -> [0, 1] -> [1, 1] (2 operations). Double all the elements: [1, 1] -> [2, 2] (1 operation). Total of operations: 2 + 1 = 3.
Example 3:
Input: nums = [4,2,5] Output: 6 Explanation: (initial)[0,0,0] -> [1,0,0] -> [1,0,1] -> [2,0,2] -> [2,1,2] -> [4,2,4] -> [4,2,5](nums).
Example 4:
Input: nums = [3,2,2,4] Output: 7
Example 5:
Input: nums = [2,4,8,16] Output: 8
Constraints:
1 <= nums.length <= 10^5
0 <= nums[i] <= 10^9
class Solution { public int minOperations(int[] nums) { int res = 0, n = nums.length; while(!allzero(nums)) { if(!alleven(nums)) { for(int i = 0; i < n; i++) { if(nums[i] % 2 != 0) { nums[i]--; res++; } } } else { for(int i = 0; i < n; i++) { nums[i] /= 2; } res++; } } return res; } public boolean allzero(int[] nums) { for(int i : nums) { if(i != 0) return false; } return true; } public boolean alleven(int[] nums) { for(int i : nums) { if((i & 1) == 1) return false; } return true; } }
我不管,我就要brute force
只要不是allzero就循环,先判断是否alleven,再进行操作。