8. String to Integer (atoi)

class Solution:
    def myAtoi(self, str: str) -> int:
        res, MAX, MIN = 0, (1 << 31) - 1, -(1 << 31)
        str = str.lstrip()
        if not str: return 0
        sign, i = 1, 0
        if str[0] in '+-':
            sign, i = 44 - ord(str[0]), 1
        for i in range(i, len(str)):
            if res > MAX or not str[i].isdigit(): break
            res = res * 10 + int(str[i])
        res *= sign
        if res < MIN: return MIN
        return res if res <= MAX else MAX

class Solution {
public:
    int myAtoi(string str) {
        if (str.empty()) return 0;
        int n = str.length();
        int b = 0;
        while (str[b] == ' ') {
            ++b;
        }
        int sign = 1;
        if (str[b] == '+' || str[b] == '-') {
            sign = 44 - str[b];
            ++b;
        }
        long res = 0;
        for (int i = b; i < n && isdigit(str[i]) && res <= INT_MAX; ++i) {
            res = res * 10 + str[i] - '0';
        }
        res *= sign;
        if (res < INT_MIN) return INT_MIN;
        return res <= INT_MAX ? res : INT_MAX;
    }
};

class Solution {
public:
    int myAtoi(string str) {
        long res = 0;
        int n = str.length();
        int b = 0;
				// 去掉开始的空格
        while (str[b] == ' ') {
            ++b;
        }
        int sign = 1;
        if (str[b] == '-' || str[b] == '+') { // 判断符号，注意b++
            if (str[b++] == '-') sign = -1;
        }
        for (int i = b; i < n && isdigit(str[i]) && res <= INT_MAX; ++i) { // 三个条件：不能越界、必须是数字、值不能超过INT_MAX（INT_MIN可行）
            res = res * 10 + str[i] - '0';
        }
        res *= sign;
        if (res > INT_MAX) return INT_MAX;
        if (res < INT_MIN) return INT_MIN;
        return res;
    }
};

class Solution {
public:
    int myAtoi(string str) {
        if (str.empty()) return 0;
        long res = 0;
        int b = 0;
        int n = str.length();
        while (b < n && str[b] == ' ') {
            ++b;
        }
        bool ispos = true;
        if (str[b] == '-') {
            ispos = false;
            ++b;
        } else if (str[b] == '+') {
            ++b;
        }
        for (int i = b; i < n && res <= INT_MAX && '0' <= str[i] && str[i] <= '9'; ++i) {
            res = res * 10 + str[i] - '0';
        }
        return ispos ? min((long)INT_MAX, res) : max((long)INT_MIN, -res);
    }
};