算法題1—翻硬幣

題目要求：

分析： 翻硬幣是將初始狀態的字符串轉變爲目標狀態的字符串，只有兩個字符串不同才進行轉換，且每次只能轉換一次，即只能轉換相鄰的兩個字符

代碼：

package net.dc.algchallenge;

import java.util.Arrays;

public class OverturnCoin {
    public static void main(String[] args) {
        String str = "**********";
        String target = "o****o****";
        System.out.println(String.format("總共翻轉：%d次", overturnCount(str, target)));

        str = "*o**o***o***";
        target = "*o***o**o***";
        System.out.println(String.format("總共翻轉：%d次", overturnCount(str, target)));
    }

    public static int overturnCount(String str, String target) {
        char[] strArr = str.toCharArray();
        char[] targetArr = target.toCharArray();
        int count = 0;
        k:
        while (!Arrays.equals(strArr, targetArr)) {
            for (int i = 0; i < strArr.length - 1; i++) {
                if (strArr[i] != targetArr[i]) {
                    strArr[i] = targetArr[i];
                    if (strArr[i + 1] == '*') {
                        strArr[i + 1] = 'o';
                    } else if (strArr[i + 1] == 'o') {
                        strArr[i + 1] = '*';
                    }
                    count++;
                    continue k;
                }
            }
        }
        return count;
    }
}

運行結果：

如有錯誤，敬請指正