KMP算法（尝试）

最新推荐文章于 2025-04-04 21:54:16 发布

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本文介绍了一种使用KMP算法解决字符串匹配问题的方法，并通过一个具体的应用实例展示了如何实现字符串匹配及计数的功能。该程序能够高效地在主字符串中查找并统计模式字符串出现的次数。

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//
//  main.c
//  test
//
//  Created by 胡博豪 on 16/2/29.
//  Copyright © 2016年 胡博豪. All rights reserved.
//

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


int *next(const char *string) {
    int *next = (int *)malloc(sizeof(int) * 100);
    next[0] = next[1] = 0;
    int i = 0, j = 1;
    while (++j < strlen(string)) {
        if (string[i] != string[j]) {
            next[j] = 0;
            i = 0;
        } else {
            i++;
            next[j] = next[j - 1] + 1;
        }
    }
    return next;
}

int KMP(const char *ms, const char *os, const int *next) {
    int i, j = 0;
    for (i = 0; i < strlen(ms);) {
        if (ms[i] == os[j]) {
            j++;
            i++;
        } else {
            if (j == 0) {
                i++;
            }
            j = next[j];
        }
        if (j == strlen(os)) {
            return i - j + 1;
        }
    }
    return -1;
}

int count_str_in_str(char *str_m, const char *str_o) {
    int count = 0, temp = 0;
    int *n = next(str_o);
    char *str_temp = str_m;
    while (temp != -1) {
        temp = KMP(str_temp, str_o, n);
        str_temp += temp + 1;
        count++;
    }
    free(n);
    return count - 1;
}

int main(int argc, const char * argv[]) {
    int n, index;
    while(~scanf("%d", &n)) {
        char **sa = (char **)malloc(sizeof(char *) * n);
        for (index = 0; index < n; index++) {
            char *temp = (char *)malloc(sizeof(char) * 100);
            scanf("%s", temp);
            sa[index] = temp;
        }
        char data[100001];
        scanf("%s", data);
        char *re = NULL;
        int amount = 0;
        for (index = 0; index < n; index++) {
            int te = count_str_in_str(data, sa[index]);
            if (te > amount) {
                amount = te;
                re = sa[index];
            }
        }
        if (amount == 0) {
            printf("exciting\n");
        } else {
            printf("%s %d\n", re, amount);
        }
        for (index = 0; index < n; index++) {
            free(sa[index]);
        }
        free(sa);
    }
    return 0;
}