LeetCode8——String to Integer (atoi)（自己编写atoi函数）

题目：

参考解法：

I think we only need to handle four cases:

discards all leading whitespaces

sign of the number

overflow

invalid input

int myAtoi(char* str)
{int sign = 1, base = 0, i = 0;while (str[i] == ' ') { i++; }//去掉空格if (str[i] == '-' || str[i] == '+') {sign = 1 - 2 * (str[i++] == '-');//这里i进行了++，注意符号也要输出的 }//判断符号while (str[i] >= '0' && str[i] <= '9') {if (base >  INT_MAX / 10 || ((base == INT_MAX / 10) &&  (str[i] - '0' > 7))) //这个条件会在加到一定程度时才触发{if (sign == 1) return INT_MAX;elsereturn INT_MIN;}base  = 10 * base + (str[i++] - '0');}return base * sign;
}

分析：

注意这里的判断条件：

base == INT_MAX / 10 && (str[i] - '0' > 7)，最后一个为什么是这个？

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LeetCode8——String to Integer (atoi)（自己编写atoi函数）

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