数据结构-KMP算法

先解决 前缀与后缀串的最长匹配长度信息(前缀或后缀都不能取整体)。如下
位置6的前缀最长串就是abcab(不能取全部,即不能为abcabc)
位置6的后缀最长串就是bcabc(不能取全部,即不能为abcabc)

public class Code_KMP {
    public static void main(String[] args) {
        String s1 = "abcdefgjhi32jkls.dfs";
        String s2 = "s.df";
        System.out.println(getIndexOf(s1,s2));
        System.out.println(getIndexOf2(s1,s2));
    }

    public static int getIndexOf(String s1, String s2){
        if(s1 == null || s2 == null || s1.length() < 1 || s1.length() < s2.length()){
            return -1;
        }

        char[] str1 = s1.toCharArray();
        char[] str2 = s2.toCharArray();
        int x = 0;
        int y = 0;
        int[] next = getNextArray(str2);
        while(x < str1.length && y<str2.length){
            if(str1[x] == str2[y]){
                x++;
                y++;
            }else if(next[y] == -1){
                x++;
            }else{
                y = next[y];
            }
        }

        return y == str2.length ? x - y : -1;
    }

    public static int[] getNextArray(char[] match){
        if(match.length == 1){
            return new int[]{-1};
        }

        int[] nextArray = new int[match.length];
        nextArray[0] = -1;
        nextArray[1] = 0;
        nextArray[2] = (match[0] == match[1] ? 1 : 0);

        for (int i = 3; i < match.length; i++) {
            int cn = nextArray[i-1];
            while(cn != -1 && match[cn] != match[i-1]){
                 cn = nextArray[cn];
            }
            nextArray[i] = cn + 1;
        }

        return nextArray;
    }


    public static int getIndexOf2(String s1, String s2){
        if(s1 == null || s2 == null || s1.length() < 1 || s1.length() < s2.length()){
            return -1;
        }

        char[] str1 = s1.toCharArray();
        char[] str2 = s2.toCharArray();
        int x = 0;
        int y = 0;
        int[] next = getNextArray2(str2);
        while(x < str1.length && y<str2.length){
            if(str1[x] == str2[y]){
                x++;
                y++;
            }else if(next[y] == -1){
                x++;
            }else{
                y = next[y];
            }
        }

        return y == str2.length ? x - y : -1;
    }

    public static int[] getNextArray2(char[] match){
        if(match.length == 1){
            return new int[]{-1};
        }

        int[] nextArray = new int[match.length];
        nextArray[0] = -1;
        nextArray[1] = 0;

        int cn = 0;
        int i = 2;
        while(i <match.length){
            if(match[0] == match[i]){ // 匹配成功
                nextArray[i++] = ++cn;
            }else if(cn > 0){
                cn = nextArray[cn];
            }else{
                nextArray[i++] = 0;
            }
        }

        return nextArray;
    }
}