<LeetCode OJ> 8. String to Integer (atoi)

来源：互联网发布：云南网星大数据编辑：程序博客网时间：2024/05/04 13:05

8. String to Integer (atoi)

Total Accepted: 82548 Total Submissions: 626140 Difficulty: Easy

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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument,

please click the reload button to reset your code definition.

首先分析：

难点在于异常情况的处理,很难考虑全面：实际上模拟人性错误习惯，以下答案为别人的答案

1、处理输入空

2、处理数字前面有空格

3，判断符号位：先出现的就是符号位

4、处理数字前面的0

5、处理输入非法，函数输出0

6、处理溢出情况

class Solution {public:    int myAtoi(string str)    {        //1，处理空情况        int len=str.size();        if(len == 0)            return 0;                    //2，过滤掉数字前面的空格        int i=0;        while(str[i] == ' ')            i++;                //3，判断符号位        int sign = 0;        if(str[i] == '-')        {            sign = 0;             i++;        }        else if(str[i] == '+')        {                sign=1;             i++;        }                 //4，过滤掉数字前面的0        while(str[i] == '0')            i++;                int res = 0, lastRes = 0;                //5，转换字符数字，不是字符数字不要处理        int limit = INT_MAX / 10;        while(i<str.size() && str[i] >= '0' && str[i] <= '9')        {            if(limit < res)                return sign == 1 ? INT_MAX : INT_MIN;//判断是否溢出(两个数相乘溢出后，溢出的结果不一定比两个乘数小)                        lastRes = res;            res = res*10 + (int)(str[i] - '0');                        if(res < lastRes)                return sign == 1 ? INT_MAX : INT_MIN;//判断是否溢出            i++;        }                return sign == 1 ? res : -1*res;    }};

注：本博文为EbowTang原创，后续可能继续更新本文。如果转载，请务必复制本条信息！

原文地址：http://blog.csdn.net/ebowtang/article/details/50511705

原作者博客：http://blog.csdn.net/ebowtang

1 0