Leetcode - Medium - 8. String to Integer (atoi) - Javascript

Leetcode - Medium - 8. String to Integer (atoi) - Javascript

   清晨6:40       Rain Lin

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8. String to Integer (atoi)

Medium

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:

1. Read in and ignore any leading whitespace.
2. 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.
3. Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.
4. 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).
5. 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.
6. 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.

 

Example 1:

Input: s = "42"
Output: 42
Explanation: The underlined characters are what is read in, the caret is the current reader position.
Step 1: "42" (no characters read because there is no leading whitespace)
         ^
Step 2: "42" (no characters read because there is neither a '-' nor '+')
         ^
Step 3: "42" ("42" is read in)
           ^
The parsed integer is 42.
Since 42 is in the range [-231, 231 - 1], the final result is 42.

解題方向：

1. 一開始以為是單純的文字移除空白和文字後再轉成 int 的函式, 但詳細了解後並非這麼簡單

2. 這邊有分成兩階段共三步驟方式處理

- 第一階段: 累計數字

    step 1. 先利用 isPositive 紀錄是否開頭為 '-', '+', '0-9' 如果在這個步驟遇到空白外的文字即立刻回傳 0

    step 2. 當紀錄 isPositive 後就開始累計 '0-9', 遇到其他文字立即中斷 while 迴圈, 結束累計

- 第二階段 處理回傳值

    step 3. 透過文字個別轉成數字時判斷是否落在 minLimit 與 maxLimit 之間, 如不是即立刻回傳對應的上下限值即可

程式碼 ：

/**
 * @param {string} s
 * @return {number}
 */
var myAtoi = function(s) {
    let index = 0;
    let isPositive = 0;
    let resultString = '';
    const digiArray = ['1', '2', '3', '4', '5', '6', '7', '8', '9', '0'];
    
    while(index < s.length) {
        const letter = s[index];
        index++;
        
        if (isPositive === 0) {
            if (letter === ' ') {
                continue;
            } else if (letter === '-') {
                isPositive = '-1';
                continue;
            } else if (letter === '+') {
                isPositive = '1';
                continue;
            } else if (digiArray.includes(letter)) {
                isPositive = '1';
                resultString += letter;
            } else {
                return 0;
            }
            
        } else {
            if (!digiArray.includes(letter)) {
                break;
            } else {
                resultString += letter;
            }
        }        
    }
    
    
    const minLimit = -2147483648;
    const maxLimit = 2147483647;
    let result = 0;
    
    for (let i in resultString) {
        result = result * 10 + isPositive * parseInt(resultString[i]);
        
        if (result < minLimit) { return minLimit};
        if (result > maxLimit) { return maxLimit};
    }
    
    return result;
};

執行成果：