思路

每次只能翻转相邻两枚硬币，因此开始状态与目标状态的不同之处一定是偶数个
对于每两个不同之处单独进行翻转，翻转次数一定是位置之差
因此只需遍历一遍将这些差相加即可，时间复杂度O(n)

java 代码

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.IOException;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        char[] start = br.readLine().toCharArray();
        char[] end = br.readLine().toCharArray();
        int temp = -1, res = 0;
        for (int i = 0; i < start.length; i++) {
            if (start[i] != end[i]) {
                if (temp == -1) {
                    temp = i;
                } else {
                    res += i - temp;
                    temp = -1;
                }
            }
        }
        bw.write(String.valueOf(res));
        bw.flush();
        bw.close();
    }
}