标签:VID Plan 代码 lang 思路 code ISE 操作 xpl
Given a non-negative integer num, return the number of steps to reduce it to zero. If the current number is even, you have to divide it by 2, otherwise, you have to subtract 1 from it.
Example 1:
Input: num = 14
Output: 6
Explanation:
Step 1) 14 is even; divide by 2 and obtain 7.
Step 2) 7 is odd; subtract 1 and obtain 6.
Step 3) 6 is even; divide by 2 and obtain 3.
Step 4) 3 is odd; subtract 1 and obtain 2.
Step 5) 2 is even; divide by 2 and obtain 1.
Step 6) 1 is odd; subtract 1 and obtain 0.
Example 2:
Input: num = 8
Output: 4
Explanation:
Step 1) 8 is even; divide by 2 and obtain 4.
Step 2) 4 is even; divide by 2 and obtain 2.
Step 3) 2 is even; divide by 2 and obtain 1.
Step 4) 1 is odd; subtract 1 and obtain 0.
Example 3:
Input: num = 123
Output: 12
Constraints:
0 <= num <= 10^6将一个数进行若干次偶数/2、奇数-1的操作,直至其变为0,求需要的操作次数。
直接照做即可。
class Solution {
public int numberOfSteps (int num) {
int steps = 0;
while (num > 0) {
num = num % 2 == 0 ? num / 2 : num - 1;
steps++;
}
return steps;
}
}
1342. Number of Steps to Reduce a Number to Zero (E)
标签:VID Plan 代码 lang 思路 code ISE 操作 xpl
原文地址:https://www.cnblogs.com/mapoos/p/14398906.html