题解 | #最长公共前缀#C+暴力解法

看没人用C做这个题，写一下自己的思路

1. 找到最短字符串。
2. 用最短字符串和其它字符串比较，找到公共的子序列
3. 时间复杂度：N+N*M

char* Com_str12(char* s1, char* s2, int n) {
    char* s = (char*)malloc(sizeof(char) * 400);
    int i = 0, j = 0;
    if (s1[0] != s2[0]) {
        s = "";
        return s;
    }
    while (i < n) {
        if (s1[i] == s2[i]) {
            s[i] = s1[i];
        } else {
            break;
        }
        i++;
    }
    return s;

}
char* longestCommonPrefix(char** strs, int strsLen ) {
    // write code here
    //暴力解法思路，先找出最短的字符串的长度，然后截取最长看是否相等，如果不是，就减一
    int i = 0, min_len = 5001, j = 0, max = 5001, len1 = 0, g = 0;
    //char* s1 = (char*)malloc(sizeof(char) * 5000);
    char* s2 = (char*)malloc(sizeof(char) * 400);//需要提前申请地址
    char* s3 = (char*)malloc(sizeof(char) * 400);

    if (strsLen == 0 || strs == NULL) {

        return "";
    }
    if (strsLen == 1) {
        return strs[0];
    }
    for (i = 0; i < strsLen; i++) {//找最短字符串
        len1 = strlen(strs[i]);
        if ( len1 < min_len) {
            min_len = strlen(strs[i]);
           // strcpy(s1, strs[i]);
            g = i;
        }
    }
    for (i = 0; i < strsLen ; i++) {//和最短字符串比较
        if (i != g) {
            s2 = Com_str12(strs[i], strs[g], min_len);
            if (max > strlen(s2)) {
                max = strlen(s2);
                strcpy(s3,s2);
            }
        }

    }
    return s3;
}