求两段中文语句的最长公共子串c语言代码

以下是基于动态规划的C语言代码实现，时间复杂度为O(m*n)，其中m和n分别是两个中文语句的长度。

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

char* LCS(char* str1, char* str2) {
    int len1 = strlen(str1);
    int len2 = strlen(str2);
    int i, j;
    int** dp = (int**)malloc(sizeof(int*) * (len1 + 1));
    for (i = 0; i <= len1; i++) {
        dp[i] = (int*)malloc(sizeof(int) * (len2 + 1));
        memset(dp[i], 0, sizeof(int) * (len2 + 1));
    }
    int max_len = 0, max_end = 0;
    for (i = 1; i <= len1; i++) {
        for (j = 1; j <= len2; j++) {
            if (str1[i - 1] == str2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
                if (dp[i][j] > max_len) {
                    max_len = dp[i][j];
                    max_end = i - 1;
                }
            }
        }
    }
    char* result = (char*)malloc(sizeof(char) * (max_len + 1));
    strncpy(result, str1 + max_end - max_len + 1, max_len);
    result[max_len] = '\0';
    for (i = 0; i <= len1; i++) {
        free(dp[i]);
    }
    free(dp);
    return result;
}

int main() {
    char str1[] = "以下是基于动态规划的C语言代码实现";
    char str2[] = "动态规划的C语言代码实现，时间复杂度为O(m*n)";
    char* result = LCS(str1, str2);
    printf("最长公共子串为: %s\n", result);
    free(result);
    return 0;
}