leetcode 1342. Number of Steps to Reduce a Number to Zero

标签：奇数   elf   --   提高效率   public   运算   bsp   def   ISE   

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

题目大意：给定一个非负整数num,返回将其变为0的步数。如果当前的数是偶数，将其除2，否则将其减去1.

思路：直接模拟，为了提高效率，除2用位运算。

C++代码1：

 1 class Solution {
 2 public:
 3     int numberOfSteps (int num) {
 4         int cnt = 0;
 5         while (num != 0) {
 6             if (num & 1) //为奇数
 7                 num--;
 8             else
 9                 num >>= 1;
10             cnt++;
11         }
12         return cnt;
13     }
14 };

C++代码二：

一般条件下，如果一个数是奇数，我们将其减去1后，下一步肯定除2，（如(5 - 1)/2=2,用了两步），而利用位运算，可以直接右移一位，会产生两步的效果:5 >> 1 = 2.

只要整数num一直减小，在变成0之前，肯定会变成1. 当数为1时，步数会多算1.

class Solution {
public:
    int numberOfSteps (int num) {
        int cnt = 0;
        while (num != 0) {
            cnt += (num & 1) ? 2 : 1;
            num >>= 1;
        }
        return cnt - 1;
    }
};

 python3代码：

1 class Solution:
2     def numberOfSteps (self, num: int) -> int:
3         cnt = 0
4         while num != 0: 
5             num, cnt = num - 1 if num % 2 else num // 2, cnt + 1
6         return cnt

leetcode 1342. Number of Steps to Reduce a Number to Zero

标签：奇数   elf   --   提高效率   public   运算   bsp   def   ISE   

原文地址：https://www.cnblogs.com/qinduanyinghua/p/12299781.html