题解 | #合法IP#面向用例编程#较繁琐#纯C

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

int main() {
    char str[20] = {'\0'};
    scanf("%s", str); // 读入IP字符串
    int symbol[4] = {0}; // 此数组用于标记'.'出现的位置
    int symbolNum = 0;
    for (int i = 0; i < 20; i++) {
        if (str[i] == '.') {
            symbol[symbolNum] = i;
            symbolNum++;
        }
    }

    for (int i = 0; i < strlen(str); i++) { // 检查str中是否存在非数字和'.'字符的其他字符，如果有，非法，输出NO，结束
        if (str[i] != '.' && (str[i] < '0' || str[i] > '9')) {
            printf("NO");
            return 0;
        }
    }

    char str1[5] = {'\0'}; // 使用四个char数组来存储str中的四个数，后面借此判断各个数是否以0开头，如果是，非法
    char str2[5] = {'\0'};
    char str3[5] = {'\0'};
    char str4[5] = {'\0'};
    int a = 0, b = 0, c = 0, d = 0;
    for (int i = 0; i < symbol[0]; i++) {
        str1[a] = str[i];
        a++;
    }
    for (int i = symbol[0] + 1; i < symbol[1]; i++) {
        str2[b] = str[i];
        b++;
    }
    for (int i = symbol[1] + 1; i < symbol[2]; i++) {
        str3[c] = str[i];
        c++;
    }
    for (int i = symbol[2] + 1; i < strlen(str); i++) {
        str4[d] = str[i];
        d++;
    }

    if (strlen(str1) == 0 || strlen(str2) == 0 || strlen(str3) == 0 || strlen(str4) == 0) { // 用例中存在.开头以及..在中间的情况，此处处理
        printf("NO");
        return 0;
    }

    int num1 = 0, num2 = 0, num3 = 0, num4 = 0; // 将str中的四个char类型的字符串转化为int数字，用于判断IP是否非法
    int factor1 = 1, factor2 = 1, factor3 = 1, factor4 = 1;
    for (int i = strlen(str1) - 1; i >= 0; i--) {
        num1 += (str1[i] - '0') * factor1;
        factor1 *= 10;
    }
    for (int i = strlen(str2) - 1; i >= 0; i--) {
        num2 += (str2[i] - '0') * factor2;
        factor2 *= 10;
    }
    for (int i = strlen(str3) - 1; i >= 0; i--) {
        num3 += (str3[i] - '0') * factor3;
        factor3 *= 10;
    }
    for (int i = strlen(str4) - 1; i >= 0; i--) {
        num4 += (str4[i] - '0') * factor4;
        factor4 *= 10;
    }

    if ((str1[0] == '0' && str1[1] != '\0') || (str2[0] == '0' && // 若某个str以0开头，则非法
            str2[1] != '\0') || (str3[0] == '0' && str3[1] != '\0') || (str4[0] == '0' &&
                    str4[1] != '\0') || strlen(str4) == 0) {
        printf("NO");
    } else if (num1 > 255 || num2 > 255 || num3 > 255 || num4 > 255) { // 若某个num大于255，则非法
        printf("NO");
    } else {
        printf("YES");
    }


    return 0;
}