题解 | #最长公共子串#

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * longest common substring
 * @param str1 string字符串 the string
 * @param str2 string字符串 the string
 * @return string字符串
 */

#include <stdio.h>
char* LCS(char* str1, char* str2 ) {
    // write code here
    int l1 = strlen(str1);
    int l2 = strlen(str2);
    int** dp = (int**)malloc(sizeof(int*) * l1);
    int max = 0;
    int endPosition = 0;
    for(int i = 0; i < l1; i++) {
        dp[i] = (int*)malloc(sizeof(int) * l2);
        
        for(int j = 0; j < l2; j++) {
            if(str1[i] == str2[j]) {
                if(i == 0 || j == 0) dp[i][j] = 1;
                else {
                    dp[i][j] = dp[i - 1][j - 1] + 1;
                    if(dp[i][j] > max) {
                        max = dp[i][j];
                        endPosition = i;
                    }
                }
            } else {
                dp[i][j] = 0;
            }
        }
    }
    char* result = (char*)malloc(sizeof(char) * max);
    if(result == NULL) {
        perror("内存分配失败");
        exit(1);
    }
    strncpy(result, str1 + endPosition - max + 1, max);
    result[max] = '\0';
    for(int i = 0; i < l1; i++) {
        free(dp[i]);
    }
    free(dp);
    return result;
}