Leetcode 每日刷题--罗马数字转整数

问题描述：给定罗马数字（字符表示）转整数

结题思路:

1. 字符串遍历，转为对应整数，由于一一对应，使用haspmap 建立hasp 表

2. 对于4，9这类特殊数字，做特殊处理，根据下图规则可以看出，

表示4，9时，前一个罗马字符小于后一个，据其处理即可，

答案C语言版本:

#include "errno.h"
#include "string.h"

#define ARRAY_SIZE(x) sizeof(x) / sizeof(x[0])

struct hash_map {
    char key;
    uint32_t value;
};

static struct hash_map roman_to_inter_convetor[] = {
    {
        .key = 'I',
        .value = 1,
    },
    {
        .key = 'V',
        .value = 5,
    },
    {
        .key = 'X',
        .value = 10,
    },
    {
        .key = 'L',
        .value =  50,
    },
    {
        .key = 'C',
        .value = 100,
    },
    {
        .key = 'D',
        .value = 500,
    },
    {
        .key = 'M',
        .value =1000,
    },
};

static struct hash_map* get_hash_map_by_key(char key) {
    uint32_t i = 0;

    for (i = 0; i < ARRAY_SIZE(roman_to_inter_convetor); i++)
        if (roman_to_inter_convetor[i].key == key)
            return &roman_to_inter_convetor[i];

    return NULL;
}

int romanToInt(char * s) {
    uint32_t i = 0;
    struct hash_map *mapped = NULL;
    uint32_t mapped_value[15] = {0};
    uint32_t result = 0;

    for (i = 0; i < strlen(s); i++) {
        mapped = get_hash_map_by_key(s[i]);
        if (!mapped)
            return -EINVAL;

        mapped_value[i] = mapped->value;
    }

    for (i = 0; i < strlen(s) -1; i++) {
        if (mapped_value[i] < mapped_value[i + 1])
            result -= mapped_value[i];
        else
            result += mapped_value[i];
    }

    result += mapped_value[i];
    return result;
}