KMP算法和BM算法的java简单实现

//参考http://www.cnblogs.com/linbingdong/p/6479537.html
//http://blog.csdn.net/henuyx/article/details/44653799
//http://www.cnblogs.com/tangzhengyue/p/4315393.html
package string;

public class KMPSearch {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        String txt = "yuchengxin is good man";
        String pat = "good";
        int[] next = new int[pat.length()];
        getNext(pat, next);
        System.out.println(Search(txt, pat, next));
    }

    //next数组为pat跳转记录状态的数组,此处是关键,注意理解
    public static int[] getNext(String pat, int[] next) {
        int N = pat.length();
        next[0] = -1;//初始条件next[0]=-1,next[1]=0
        int k = -1;
        int j = 0;
        while (j < N - 1) {
            if (k == -1 || pat.charAt(j) == pat.charAt(k)) {
                next[++j] = ++k; //若相等,则j的下一个跳到k的下一个处
                if (pat.charAt(j) == pat.charAt(k)) {
                    next[j] = next[k];//若继续相等,则再往前跳
                }
            } else {
                k = next[k];//不相等则移回到k处
            }
        }
        return next;
    }

    public static int Search(String txt, String pat, int[] next) {
        int M = txt.length();
        int N = pat.length();
        int i = 0, j = 0;
        while (i < M && j < N) {
            if (j == -1 || txt.charAt(i) == pat.charAt(j)) {
                j++;
                i++;
            } else {
                j = next[j];
            }
        }
        if (j == N)
            return i - j;
        else
            return -1;
    }
}
//参考算法第四版代码
package string;

public class BoyerMooreSearch {
    //注意此处right[]的构造
    public static void getRight(String pat, int[] right) {
        for (int i = 0; i < 256; i++) {
            right[i] = -1;
        }
        for (int j = 0; j < pat.length(); j++) {
            right[pat.charAt(j)] = j;
        }
    }

    public static int Search(String txt, String pat, int[] right) {
        int M = txt.length();
        int N = pat.length();
        int skip;
        for (int i = 0; i < M - N; i += skip) {
            skip = 0;
            for (int j = N - 1; j >= 0; j--) {
                if (pat.charAt(j) != txt.charAt(i + j)) {
                    skip = j - right[txt.charAt(i + j)];
                    if (skip < 1)
                        skip = 1;
                    break;
                }
            }
            if (skip == 0)
                return i;
        }
        return -1;
    }

    public static void main(String[] args) {
        String txt = "THIS IS A BIG TIGER";
        String pat = "IG";
        int[] right = new int[256];
        getRight(pat, right);
        System.out.println(Search(txt, pat, right));
    }

}

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