Question

你好我尝试在一个字符串数组中打印通过以下方式：

输入：big =“12xy34”，small =“xy”输出：“** xy **”
输入：big =“”12xt34“”，small =“xy”输出：“******”
输入：big =“12xy34”，small =“1”输出：“1 *****”
输入：big =“12xy34xyabcxy”，small =“xy”输出：“** xy ** xy *** xy”
输入：big =“78abcd78cd”，small =“78”输出：“78 **** 78 **”

我需要写一个条件来接收吗？



 public static String stars(String big, String small) {
      //throw new RuntimeException("not implemented yet ");
     char[] arr = big.toCharArray();
    for (int i = 0; i < arr.length; i++) {
        if (big.contains(small) ) {
            arr[i] = '*';
        }
    }
        String a = Arrays.toString(arr);
     return big+""+a;

public static String stars(String big, String small) { //throw new RuntimeException("not implemented yet "); char[] arr = big.toCharArray(); for (int i = 0; i < arr.length; i++) { if (big.contains(small) ) { arr[i] = '*'; } } String a = Arrays.toString(arr); return big+""+a;

Answer 1

<强>算法：

分别将big和small String's转换为char[]数组＆＃39; bigC和smallC
迭代big String
在迭代期间的每个索引处，确定是否存在可能的开始当前字符的子字符串
如果存在子字符串的可能性，请将big字符串迭代中的索引推进small String
否则，请用*

<强>代码：

public class StringRetainer {

    public static void main(String args[]) {
        String big[] = {"12xy34", "12xt34", "12xy34", "12xy34xyabcxy", "78abcd78cd"};
        String small[] = {"xy", "xy", "1", "xy", "78"};
        for(int i = 0; i < big.length & i < small.length; i++) {
            System.out.println("Input: big = \"" + big[i] + "\", small = \"" + small[i] + "\" output : \"" + stars(big[i], small[i]) + "\"");
        }
    }

    public static String stars(String big, String small) {
        //String to char[] array conversions
        char[] bigC = big.toCharArray();
        char[] smallC = small.toCharArray();
        //iterate through every character of big String and selectively replace
        for(int i = 0; i < bigC.length; i++) {
            //flag to determine whether small String occurs in big String
            boolean possibleSubString = true;
            int j = 0;
            //iterate through every character of small String to determine
            //the possibility of character replacement
            for(; j < smallC.length && (i+j) < bigC.length; j++) {
                //if there is a mismatch of at least one character in big String
                if(bigC[i+j] != smallC[j]) {
                    //set the flag indicating sub string is not possible and break
                    possibleSubString = false;
                    break;
                }
            }
            //if small String is part of big String,
            //advance the loop index with length of small String
            //replace with '*' otherwise
            if(possibleSubString)
                i = i+j-1;
            else
                bigC[i] = '*';
        }
        big = String.copyValueOf(bigC);
        return big;
    }

}

注意：

这是一种可能的解决方案（传统的做法）

使用内置String / StringBuffer / StringBuilder方法
看起来没有直接的方法来实现这一目标

如何保留匹配的子字符串并替换Java String中不匹配的子字符串

1 个答案: