Leetcode 833:字符串替换取决于索引编号

Leetcode 833: String replacement depend upon index numbering

这个问题是我上一个问题的一部分。

示例 1: 输入:S = "abcd", indexes = [0,2], sources = ["a","cd"], targets = ["eee","ffff"] 输出:eeebffff 解释:a 从 S 中的索引 0 开始,因此它被 eee 替换。 cd 从 S 中的索引 2 开始,因此它被替换为 ffff

示例 2: 输入:S = "abcd", indexes = [0,2], sources = ["ab","ec"], targets = ["eee","ffff"] 输出:"eeecd" 解释:"ab" 从 S 中的索引 0 开始,因此它被 "eee" 取代。 "ec" 不是从原始 S 中的索引 2 开始,所以我们什么都不做。

public class Q833 {
public static void main(String args[]){
    String S="abcd";
    int[] indexes  = {0, 2};
    String[]sources={"ab","cd"};
    String[] targets = {"eee", "ffff"};
    Solve833 ob833=new Solve833();
    System.out.println(ob833.findReplaceString(S,indexes,sources,targets));
}
}
 class Solve833{
 public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
    char[] array = S.toCharArray();
    StringBuilder result = new StringBuilder();
    int counter = 0;
    String s = "";
    for (String n:sources)
        s+= n;
    char[] c = s.toCharArray();

    for (int i = 0; i < array.length; i++) {
        if(array[indexes[counter]]==c[counter]){
            result.append(targets[counter]);
            if(counter<=indexes.length) {
                counter++;
            }

        }
        else
            result.append(array[i]);
    }


    return result.toString();
}

}

代码输出:第一个例子

预期输出:Output: "eeebffff"

我的输出:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 2 at Leetcode.Solve833.findReplaceString(Q833.java:30) at Leetcode.Q833.main(Q833.java:16)

代码Output:2nd示例

预期输出:"eeecd"

我的输出:eeebcd。所以这里 a b 不见了。我该如何处理?

您可以像这样更改方法来打印结果,

public class Q833 {
    public static void main(String args[]) {
        String S = "abcd";
        int[] indexes = {0, 2};
        String[] sources = {"a", "cd"};
        String[] targets = {"eee", "ffff"};
        Solve833 ob833 = new Solve833();
        System.out.println(ob833.findReplaceString(S, indexes, sources, targets));
    }
}

class Solve833 {
    public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        StringBuffer result = new StringBuffer(S);
        for (int i = 0; i < indexes.length; i++) {
            if (sources[i].equals(result.substring(indexes[i], indexes[i] + sources[i].length()))) {
                result.replace(indexes[i], indexes[i] + sources[i].length(), targets[i]);
                if (i < indexes.length - 1)
                    indexes[i + 1] = indexes[i + 1] + targets[i].length() - sources[i].length();
            }
        }
        return result.toString();
    }
}

你的问题是你不应该 array[indexes[counter]]==c[counter] 来确定 i-th 源字符串是否出现在索引 iS 中。您的判断只检查源字符串的第一个字符。

这个问题的关键是我们如何才能正确找到索引,因为当我们试图得到结果时,索引(在何处用目标字符串替换源字符串)可能会改变。

试试这个代码:

  public String findReplaceString(String S, int[] indexes, String[] sources, String[] targets) {
        StringBuilder sb=new StringBuilder(S);
        int[] offsets=new int[indexes.length];

        for(int i=0;i<indexes.length;i++){

            if(S.substring(indexes[i],indexes[i]+sources[i].length()).equals(sources[i])){
                int offset=0;
                for(int j=0;j<i;j++){
                    if(indexes[j]<indexes[i])
                    offset+=offsets[j];
                }
                sb.replace(indexes[i]+offset,indexes[i]+sources[i].length()+offset,targets[i]);
                offsets[i]=targets[i].length()-sources[i].length();

            }
        }

        return sb.toString();
    }