8. String to Integer (atoi)

Description

Implement atoi which converts a string to an integer.

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.

Note:

  • Only the space character ' ' is considered as whitespace character.
  • Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−231, 231 − 1]. If the numerical value is out of the range of representable values, INT_MAX (231 − 1) or INT_MIN (−231) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

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.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−2^31) is returned.

Tags: Math, String

题意

给你两个二进制串,求其和的二进制串。

题解

思路1

按照小学算数那么来做,用 carry 表示进位,从后往前算,依次往前,每算出一位就插入到最前面即可,直到把两个二进制串都遍历完即可。

package Solution

import (
    "math"
    "strconv"
)

func myAtoi(str string) int {
    var i, num int
    sign := 1
    if str == "" {
        return 0
    }
    //    去除前面空格
    for i < len(str) && str[i] == 32 {
        i++
    }

    if i >= len(str) {
        return 0
    }

    //判断符号
    //    +
    if str[i] == 43 {
        i++
    //    -
    } else if str[i] == 45 {
        sign = -1
        i++
    }
    //    依次将每个字符转换为数字
    for ; i < len(str); i++ {
        //    后面为字符时,直接返回前面的数字
        if str[i] != 0 && (str[i] < 48 || str[i] > 57) {
            return num * sign
        }
        //    获得每次计算的结果
        n, _ := strconv.Atoi(string(str[i]))
        num = num*10 + n
        //    检查越界【越界就返回极限值】
        if num*sign < math.MinInt32 {
            return math.MinInt32
        } else if num*sign > math.MaxInt32 {
            return math.MaxInt32
        }
    }
    return num * sign
}

思路2

思路2 ```go

```

结语

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