LeetCode_8: String to Integer (atoi)

# 引言

题目链接：https://leetcode.com/problems/string-to-integer-atoi/description/

# 题目大意

实现一个atoi函数。输入一串字符串, 扫描字符串, 跳过前面的空格, 直到遇上数字或正负符号才开始做转换, 而再遇到非数字或字符串结束时('\0')才结束转换, 并将结果返回。

• Example

Input: "42"
Output: 42

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical
             digit or a +/- sign. Therefore no valid conversion could be performed.

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−231) is returned.

# 题解

# 一句话题解

按照题意模拟流程。遇上第一个非空字符判断是否为正负号并标记, 后续继续按位来计算结果, 当遇到非数字字符, 返回当前结果, 并加上符号标记(同时处理下0值返回和数据溢出即可)。

复杂度 O(n)

# AC代码

c++版本

class Solution
{
  public:
    int myAtoi(string str)
    {
        int len = str.length();
        if (len < 1)
        {
            return 0;
        }
        int retSymbol = 1;
        int retNum = 0;
        int index = 0;
        while (index < len && str[index] == ' ')
        {
            ++index;
        }
        if (index < len && (str[index] == '-' || str[index] == '+'))
        {
            retSymbol = 1 - 2 * ('-' == str[index++]);
        }
        for (; index < len; ++index)
        {
            if (str[index] < '0' || str[index] > '9')
            {
                break;
            }
            if ((retNum > INT_MAX / 10) || (retNum == INT_MAX / 10 && (str[index] - '0') > INT_MAX % 10))
            {
                return 1 == retSymbol ? INT_MAX : INT_MIN;
            }
            retNum = retNum * 10 + str[index] - '0';
        }
        return retSymbol * retNum;
    }
};

go版本

const (
	INT32_MAX = int(^uint32(0) >> 1)
	INT32_MIN = ^INT32_MAX
)

func myAtoi(str string) int {
	lens := len(str)
	if lens < 1 {
		return 0
	}
	retSymbol, retNum, index := 1, 0, 0
	for index < lens && str[index] == ' ' {
		index++
	}
	if index < lens && str[index] == '+' {
		index++
	} else if index < lens && str[index] == '-' {
		retSymbol = -1
		index++
	}
	for ; index < lens; index++ {
		if str[index] < '0' || str[index] > '9' {
			break
		}
		if retNum > INT32_MAX/10 || (retNum == INT32_MAX/10 && int(str[index]-'0') > INT32_MAX%10) {
			if 1 == retSymbol {
				return INT32_MAX
			}
			return INT32_MIN
		}
		retNum = retNum*10 + int(str[index]-'0')
	}
	return retNum * retSymbol
}