LC题解-重复的子字符串-KML算法

解题思路

  • KML算法，计算最长相当前后缀长度，字符串长度减最长相当前后缀长度即为最小的重复字符串，字符串长度余最小重复字符串长度为0则说明字符串为重复字符串
  • 将字符串相加，同时掐头去尾2个，如果新字符串包含原字符串，则说明字符串是重复字符串

KML 解题源码

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public boolean repeatedSubstringPattern(String s) {
    if (s == null || s == "" || s.length() == 1) {
        return false;
    }
    int[] next = getNext(s);
    int preNextLen = next[next.length - 1];
    return preNextLen > 0 && next.length % (next.length - preNextLen) == 0;
}

private int[] getNext(String s) {
    char[] sc = s.toCharArray();
    int j = 0;

    int[] next = new int[s.length()];

    for (int i = 1; i < sc.length; i++) {
        while (j > 0 && sc[j] != sc[i]) {
            j = next[j - 1];
        }
        if (sc[j] == sc[i]) {
            j++;
        }
        next[i] = j;
    }

    return next;
}

相加解题源码

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public boolean repeatedSubstringPattern(String s) {
    if(s == null || s.length() < 2) {
        return false;
    }
    String ns = s + s;
    return strStr(ns.substring(1, ns.length() - 1), s) >= 0;
}

private int strStr(String ns, String s) {
    int[] next = getNext(s);

    char[] nsc = ns.toCharArray();
    char[] sc = s.toCharArray();

    int j = 0;

    for(int i = 0; i < nsc.length; i++) {
        while(j > 0 && nsc[i] != sc[j]) {
            j = next[j - 1];
        }

        if(nsc[i] == sc[j]) {
            j++;
            if(j == sc.length){
                return i - (j - 1);
            }
        }
    }

    return -1;
}

private int[] getNext(String s) {
    int[] next = new int[s.length()];

    int j = 0;
    char[] cs = s.toCharArray();
    for(int i = 1; i < s.length(); i++) {
        while(j > 0 && cs[i] != cs[j]) {
            j = next[j - 1];
        }
        
        if(cs[j] == cs[i]) {
            j++;
        }

        next[i] = j;
    }

    return next;
}