8. String to Integer(Atoi)

题目:

解答:

关键是注意怎么判定溢出的问题,测试用例一定要保证有INT_MAX, INT_MIN, INT_MAX + 1, INT_MIN - 1;

当然,如果result设置为long long就完全没问题了,但是这样做就太没有难度了

代码:

class Solution {
public:
    int myAtoi(string str) {
        int result = 0, len = str.size(), sign = 1;
        int index = 0;
        while(index < len && str[index] == ' ')    ++index;
        if(index < len && (str[index] == '+' || str[index] == '-'))
        {
            sign = str[index] == '+' ? 1 : -1;
            ++index;
        }
        while(index < len){
            if(str[index] > '9' || str[index] < '0')
                break;
            int remain = str[index] - '0';
            if(sign > 0 && (result > INT_MAX / 10 || result == INT_MAX / 10 && remain > 7))
                return INT_MAX;
            int mintmp = INT_MIN / 10 * -1;
            if(sign < 0 && (result > mintmp || result == mintmp && remain > 8))
                return INT_MIN;
            result = result * 10 + remain;
            ++index;
        }
        return sign ? result * sign : result;
    }
};

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转载自blog.csdn.net/u013434984/article/details/83894266