0008. String to Integer (atoi) (M)

String to Integer (atoi) (M)

题目

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: \([−2^{31}, 2^{31} − 1]\). If the numerical value is out of the range of representable values, INT_MAX \((2^{31} − 1)\) or INT_MIN \((−2^{31})\) 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.

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题意

给定一个字符串，要求返回除去空白字符后开头形成的第一个整数(包括+/-)。如果开头无法形成整数，则返回0；如果形成的整数溢出，返回int的上/下界。

思路

按部就班走即可，主要注意判断溢出的方法。

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代码实现

Java

class Solution {
    public int myAtoi(String str) {
        int sum = 0;
        boolean isMinus = false;	// 记录正负
        str = str.trim();			// 去空白符
        // 空字串或开头不为'+'、'-'或数字时返回
        if (str.length() == 0 || !(str.charAt(0) == '-' || str.charAt(0) == '+' || isInt(str.charAt(0)))) {
            return 0;
        }
        // 记录正负
        if (str.charAt(0) == '-') {
            isMinus = true;
        }
    	// 第一个字符就是数字则先累加
        if (isInt(str.charAt(0))) {
            sum += str.charAt(0) - '0';
        }

        for (int i = 1; i < str.length(); i++) {
            // 当前字符为数字时，需要进行累加
            if (isInt(str.charAt(i))) {
                // 判断是否向上溢出
                if (!isMinus &&
                        ((sum == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7) || sum > Integer.MAX_VALUE / 10)) {
                    return Integer.MAX_VALUE;
                }
                // 判断是否向下溢出
                if (isMinus &&
                        ((sum == Integer.MIN_VALUE / 10 && str.charAt(i) - '0' > 8) || sum < Integer.MIN_VALUE / 10)) {
                    return Integer.MIN_VALUE;
                }
                sum = isMinus ? sum * 10 - (str.charAt(i) - '0') : sum * 10 + (str.charAt(i) - '0');
            } else {
                break;
            }
        }
        return sum;
    }

    private boolean isInt(char c) {
        return (c >= '0' && c <= '9');
    }
}

JavaScript

Api实现

/**
 * @param {string} str
 * @return {number}
 */
var myAtoi = function(str) {
    let res = Number.parseInt(str)
    
    if (res !== res ) return 0																	// 判断NaN
    if (res < -Math.pow(2, 31)) return -Math.pow(2, 31)					// 判断下界
    if (res > Math.pow(2, 31) - 1) return Math.pow(2, 31) - 1		// 判断上界
    
    return res
};

正常实现

/**
 * @param {string} str
 * @return {number}
 */
var myAtoi = function (str) {
  str = str.trim()
  let isMinus = false
  let res = 0
  let maxValue = Math.pow(2, 31) - 1
  let minValue = Math.pow(2, 31)

  for (let i = 0; i < str.length; i++) {
    if (isNumber(str[i])) {
      let num = Number(str[i])
      
      // 溢出判断
      if (!isMinus && (res > Math.trunc(maxValue / 10) || (res === Math.trunc(maxValue / 10) && num > 7))) {
        return maxValue
      }
      if (isMinus && (res > Math.trunc(minValue / 10) || (res === Math.trunc(minValue / 10) && num > 8))) {
        return -minValue
      }
      
      res = res * 10 + num
    } else if (i === 0 && (str[i] === '-' || str[i] === '+')) {
      isMinus = str[i] === '-'
    } else {
      break
    }
  }

  return res * (isMinus ? -1 : 1)
}

let isNumber = function (c) {
  if (c === ' ') return false				// 注意空白字符串转型后会变为0
  return !Number.isNaN(Number(c))
}