String to Integer (atoi) (M)

題目

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.

題意

給定一個字串，要求回傳除去空白字符后開頭形成的第一個整數(包括+/-)，如果開頭無法形成整數，則回傳0；如果形成的整數溢位，回傳int的上/下界，

思路

按部就班走即可，主要注意判斷溢位的方法，

代碼實作

Java

class Solution {
    public int myAtoi(String str) {
        int sum = 0;
        boolean isMinus = false;	// 記錄正負
        str = str.trim();			// 去空白符
        // 空字串或開頭不為'+'、'-'或數字時回傳
        if (str.length() == 0 || !(str.charAt(0) == '-' || str.charAt(0) == '+' || isInt(str.charAt(0)))) {
            return 0;
        }
        // 記錄正負
        if (str.charAt(0) == '-') {
            isMinus = true;
        }
    	// 第一個字符就是數字則先累加
        if (isInt(str.charAt(0))) {
            sum += str.charAt(0) - '0';
        }

        for (int i = 1; i < str.length(); i++) {
            // 當前字符為數字時，需要進行累加
            if (isInt(str.charAt(i))) {
                // 判斷是否向上溢位
                if (!isMinus &&
                        ((sum == Integer.MAX_VALUE / 10 && str.charAt(i) - '0' > 7) || sum > Integer.MAX_VALUE / 10)) {
                    return Integer.MAX_VALUE;
                }
                // 判斷是否向下溢位
                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實作

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

正常實作

/**
 * @param {string} str
 * @return {number}
 */
var myAtoi = function (str) {
  str = str.trim()
  let isMinus = false
  let res = 0
  let maxValue = https://www.cnblogs.com/mapoos/p/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])
      
      // 溢位判斷
      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				// 注意空白字串轉型后會變為0
  return !Number.isNaN(Number(c))
}

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