Given the binary representation of an integer as a string s, 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 is guaranteed that you can always reach one for all test cases.
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 <= 500s consists of characters '0' or '1's[0] == '1'class Solution {
public int numSteps(String s) {
char[] S = s.toCharArray();
int N = S.length, cy = 0, ct = 0;
for (int i = N - 1; i > 0; i--) {
if (cy == 1) {
if (S[i] == '0') {
S[i] = '1';
cy = 0;
} else
S[i] = '0';
}
if (S[i] == '1') {
S[i] = '0';
cy = 1;
ct += 2;
} else
ct++;
}
if (cy == 1){
if(S[0]=='1')
ct++;
else ct+=2;
}
return ct;
}
}