LeetCode解题记录（8）

LeetCode解题记录（8）

目录

LeetCode解题记录8
目录

前言

正文
题目

解法一

前言

我将慢慢开始做LeetCode上的题，并做解题记录发布在这里。我每题会给出一到多个解法，记录思考过程。我算法巨烂，是想通过这种方式稍微补补，基本功和我一样差的小伙伴可以和我一起共勉，有大神路过可以指点一二，我感激不尽。解题的最底要求是能通过LeetCode的检测，我不会丧病的为了各种提高效率在一个题上纠缠不休，所以最终解法可能也不怎么样，大家看看就好~编程语言一律采用C#。题解代码会按编号发布到我的GitHub上（仅保留最终解法）。

正文

题目

problem008：

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.

Requirements for atoi: The function first discards as many whitespace

characters as necessary until the first non-whitespace character is

found. Then, starting from this character, takes an optional initial

plus or minus sign followed by as many numerical digits as possible,

and interprets them as a numerical value.

The string can contain additional characters after those that form the

integral number, which are ignored and have no effect on the behavior

of this function.

If the first sequence of non-whitespace characters in str is not a

valid integral number, or if no such sequence exists because either

str is empty or it contains only whitespace characters, no conversion

is performed.

If no valid conversion could be performed, a zero value is returned.

If the correct value is out of the range of representable values,

INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

解法一

public class Solution {
public int MyAtoi(string str) {
int index = 0;
int len = str.Length;
int flag = 1;
long num = 0;
long flag_num = 0;
if(len > 0)
{
while(str[index] == ' ')
{
index++;
}
if(str[index] == '-')
{
flag = -1;
}
if(str[index] == '-' || str[index] == '+')
{
if(str[index+1] == '-' || str[index+1] == '+')
{
return 0;
}
index++;
}
while(index < len && (str[index] >='0') && (str[index] <= '9'))
{
num = num * 10 + str[index] - '0';
index++;
flag_num = flag * num;
if(flag_num >= Int32.MaxValue)
{
flag_num = Int32.MaxValue;
break;
}
if(flag_num <= Int32.MinValue)
{
flag_num = Int32.MinValue;
break;
}
}
}
return (int)flag_num;
}
}

这个题难在对多种情况的考虑。效率为71ms。