Flip a coin (greedy)

Problem Description

Xiao Ming is playing a "coin flip" game.

There are several coins arranged in a row on the table. We use * for heads and o for tails (lowercase, not zero).

For example, the possible situation is: **oo***oooo

If you flip the two coins on the left at the same time, it becomes: oooo***oooo

Now Xiaoming's question is: If the initial state and the target state to be reached are known, and only two adjacent coins can be flipped at the same time at a time, what is the minimum number of flips for a specific situation?

We agree that flipping two adjacent coins is called a one-step operation, then the requirement:
input format

Two lines of equal-length strings represent the initial state and the target state to be achieved, respectively. length of each line < 1000
output format

An integer representing the minimum number of operation steps.
Sample input 1

---

o****o****
Sample output 1
5Sample
input 2
o**o***o**
o***o**o**
Sample output 2
1

Solution:
Ideas: Contrast successively, if there is a difference, flip the last digit, compare sequentially, and record the number of flips.
code show as below:

import java.util.Scanner;

/** 
 * @author 作者 : Cactus
 * @version 创建时间：2018-3-27 下午07:58:17 
 */
public class Main {
    private static int count = 0, length;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String strStart = sc.nextLine();
        String strEnd = sc.nextLine();
        sc.close();
        char[] arrStart = strStart.toCharArray();
        char[] arrEnd = strEnd.toCharArray();
        length = strStart.length();
        for(int i = 0; i < length; i++){
            if(arrStart[i] != arrEnd[i]){
                count++;
                arrStart[i + 1] = flipSingle(arrStart[i + 1]);
            }
        }
        System.out.println(count);
    }
    private static char flipSingle(char x){
        if('o' == x){
            return '*';
        }else{
            return 'o';
        }
    }
}