String to Integer (atoi)

一. String to Integer (atoi)

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

Difficulty:Medium

TIME：18MIN

解法

這道題的的意思是說把字符串轉換爲int類型的數字，轉換的過程要滿足一下幾個要點：

• 遇到有效字符（+,-,0-9）前的空白符直接跳過
• 遇到非空白符非有效字符那麼直接輸出已經構造的數字（比如遇到字母）
• 遇到有效字符後除非遇到數字，不然直接輸出已經構造的數字（比如-+2，輸出0，+2-4輸出2）

int myAtoi(string str) {
    long result = 0;
    int sign = 1;
    bool show = false;  //遇到有效字符的標記，一但遇到有效字符就將該標記置爲true
    for (int i = 0; i < str.size(); i++) {
        if ((str[i] - '0') >= 0 && (str[i] - '0') <= 9) {
            result = 10 * result + str[i] - '0';
            show = true;
        }
        else if (show) //遇到有效字符後，除非是數字，不然就break
            break;
        else if (str[i] == '-') {
            sign = -1;
            show = true;
        }
        else if (str[i] == '+')
            show = true;
        else if (str[i] == '\n' || str[i] == '\r' || str[i] == ' ')
            continue;
        else  //其餘情況一律break
            break;
        if (result * sign > INT32_MAX)
            return INT32_MAX;
        else if (result * sign < INT32_MIN)
            return INT32_MIN;
    }
    return (int)(result * sign);
}

代碼的時間複雜度爲O(n) 。

總結

這道題並沒有考慮8進制和16進制的情況，不過其實處理起來應該差不了多少。