【leetcode】1404. Number of Steps to Reduce a Number in Binary Representation to One

题目如下：

Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:

• If the current number is even, you have to divide it by 2.

• If the current number is odd, you have to add 1 to it.

It's guaranteed that you can always reach to one for all testcases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.  

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.  

Example 3:

Input: s = "1"
Output: 0

Constraints:

• 1 <= s.length <= 500
• s consists of characters '0' or '1'
• s[0] == '1'

解题思路：依次判断s的最后一位。如果是0，表示为偶数，去掉最后的0即可；如果是1，表示为奇数，做加法操作。

代码如下：

class Solution(object):
    def numSteps(self, s):
        """
        :type s: str
        :rtype: int
        """
        res = 0
        l = list(s)
        while len(l) > 1:
            v = l.pop(-1)
            res += 1
            if int(v) == 0:
                continue
            else:
                l.append('0')
                for i in range(len(l)-2,-1,-1):
                    if l[i] == '0':
                        l[i] = '1'
                        break
                    else: l[i] = '0'
                if l[0] == '0':l = ['1'] + l

        return res