LeetCode 8-9 Solutions

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本文提供了一个C++实现的StringtoInteger(atoi)函数和判断整数是否为回文数的方法。atoi函数能够正确处理字符串中的空格、符号，并有效避免整数溢出。回文数判断通过反转整数的后半部分来实现，但存在一定的整数反转溢出风险。

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8. String to Integer(atoi)

class Solution {
public:
    bool isDigit(char c) { return c >= '0' and c <='9'; }

    int myAtoi(string str) {
        size_t idx = 0;
        size_t len = str.length();
        int sign = 1;
        double val = 0;

        for (size_t i = 0; i < len; ++i) { // 1st none white space character
            if (str[i] == ' ') ++idx;
            else break;
        }
        // while (str[idx] == ' ') ++idx;
        if (idx == len) return 0;

        // sign(followed by integeral characters)
        if (str[idx] == '+') {
            sign = 1;
            if (!isDigit(str[++idx])) {
                return 0;
            }
        } else if (str[idx] == '-') {
            sign = -1;
            if (!isDigit(str[++idx])) {
                return 0;
            }
        }

        for (; idx < len; ++idx) {
            if (!isDigit(str[idx])) break;
            val = (str[idx] - '0') + val * 10;
        }

        val *= sign;
        if (val > INT_MAX) return INT_MAX;
        else if (val < INT_MIN) return INT_MIN;
        else return val;
    }
};

this code can pass all the testcases, but has ugly style. One should pay attention on the type of val, long long or unsigned long long will fail on some testcases. And I still haven’t figured out the reason.

9. Palindrome Number

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0) return false;
        int _x = x, y = 0;
        while (x) {
            y = y * 10 + x % 10;
            x /= 10;
        }
        return _x == y;
    }
};

this solution can pass all the testcases, but may has problem on (reversed)integer overflow(say input number is 1463847418), and used two extra space.