Leetcode题目 String to Integer (atoi)

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概述

原题目地址
Implement the myAtoi(string s) function, which converts a string to a 32-bit signed integer (similar to C/C++'s atoi function).

The algorithm for myAtoi(string s) is as follows:

Read in and ignore any leading whitespace.
Check if the next character (if not already at the end of the string) is ‘-’ or ‘+’. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.
Read in next the characters until the next non-digit charcter or the end of the input is reached. The rest of the string is ignored.
Convert these digits into an integer (i.e. “123” -> 123, “0032” -> 32). If no digits were read, then the integer is 0. Change the sign as necessary (from step 2).
If the integer is out of the 32-bit signed integer range [-231, 231 - 1], then clamp the integer so that it remains in the range. Specifically, integers less than -231 should be clamped to -231, and integers greater than 231 - 1 should be clamped to 231 - 1.
Return the integer as the final result.
Note:

Only the space character ’ ’ is considered a whitespace character.
Do not ignore any characters other than the leading whitespace or the rest of the string after the digits.

简单来说就是要求实现atoi()的作用，参数为string类型。小白太菜，用暴力枚举法调了很久才过。

#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<algorithm>
using namespace std;

int myAtoi(string s)
{
	int sign = 1;
	bool signflag = 0;
	bool zeroflag = 0;//0表示有可能出现“前面的0”； 1则相反
	vector<int> st;
	for (int i = 0; i < s.size(); i++)
	{
		if (signflag == 0 && s[i] == ' ')continue;
		else if (signflag == 0 && s[i] == '+')signflag = 1;
		else if (signflag == 0 && s[i] == '-')signflag = 1, sign = -1;
		else if (s[i] == '0')//检查是否为“前面的0 ” 
		{
			if (zeroflag == 1)
			{
				signflag = 1, st.push_back(0);
			}
			else signflag = 1;//跳过 
		}
		else if (s[i] > '0' && s[i] <= '9')
			signflag = zeroflag = 1, st.push_back(s[i] - 48);//注意ASCII转换
		else
			break;//其他情况(出现字母etc)，直接跳出循环，输出已扫描到的合法部分
	}
	int n = 0;
	long long num = 0, sum = 0;//循环计次 
	while (!st.empty())
	{
		num = st.back();
		st.pop_back();
		num *= pow(10, n);
		sum += num;
		n++;
		if (n > 10)//防止数字过大，中途num溢出
		{
			if (sign == 1)return  pow(2, 31) - 1;
			else return-pow(2, 31);
		}
	}
	long long res = sign * sum;
	if (res > pow(2, 31) - 1)res = pow(2, 31) - 1;
	else if (res < -pow(2, 31))res = -pow(2, 31);
	return res;
}