[3 questions per day (2)] minimum operand for generating alternate binary strings

subject ：

Give you a character only ‘0’ and ‘1’ Composed string s . In one step , You can put any ‘0’ become ‘1’ , Or will ‘1’ become ‘0’ .

Alternate strings Defined as ： If there is no case that two adjacent characters are equal in the string , So the string is an alternate string . for example , character string “010” It's an alternate string , And strings “0100” No .

Return to make s become Alternate strings The required least Operands .

Example 1：

Input ：s = “0100”
Output ：1
explain ： If you change the last character to ‘1’ ,s It becomes “0101” , That is, it conforms to the alternate string definition .
Example 2：

Input ：s = “10”
Output ：0
explain ：s It's already an alternate string .
Example 3：

Input ：s = “1111”
Output ：2
explain ： need 2 Step operation to get “0101” or “1010” .

Tips ：

1 <= s.length <= 10^4
s[i] yes ‘0’ or ‘1’

Ideas ：

Actually , Alternating strings are two cases , The first is ‘1010…’, The second is 0101…, namely
1. Even digits are 0, Odd bits are 1
In this case , The value of any bit is the same as the index parity , namely s[i]%2==i%2, If not satisfied , That is, this bit needs to be changed , Count cnt1++
2. Even digits are 1, Odd bits are 0
In this case , The value and index parity of any bit are different , namely s[i]%2!=i%2, If not satisfied , That is, this bit needs to be changed , Count cnt2++

java Code ：

class Solution {
    public int minOperations(String s) {
        int step1=0,step2=0;
        for (int i = 0; i <s.length() ; i++) {
            if((s.charAt(i)-'0')%2!=i%2){
                step1++;
            }else {
                step2++;
            }
        }
        return Math.min(step1,step2);
    }
}