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'

class Solution {
    public int numSteps(String s) {
        int n = s.length(), res = 0, i = n - 1, carry = 0;
        char[] arr = s.toCharArray();
        while (i > 0) {
            if (carry == 1 && arr[i] == '0' || carry == 0 && arr[i] == '1') {
                res += 2;
                carry = 1;
            }
            else if (arr[i] == '0' && carry == 0) {
                res++;
            } else {
                res++;
                carry = 1;
            }
            i--;
        }
        if (carry > 0) res++;
        return res;
    }
}

https://leetcode.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one/discuss/671937/Java-mimic-adding

public int numSteps(String s) {
        int n = s.length(), res = 0, i = n - 1, carry = 0;
        while (i > 0) {
            if (s.charAt(i--) - '0' + carry == 1) {  // curr digit + carry is '1'; need to extra adding '1' operation;
                res++;
                carry = 1;
            }
            res++;   // dividing 2;
        }
        return res + carry;  // first digit must be 1, after add carry, need extra dividing 2;
    }

从后往前整，如果当前位 + carry位 == 0则仅需要右移一位（除以2），此时res++

如果当前位 + carry位 == 1， 则先需要+1，而且carry位此时变成1，res +=2；

如果最后剩下一位那必然是1，如果此时carry位还是1，就要再右移一位，此时res++