題目：

難度：中等

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.

解釋下題目巧娱，給定一串字符串斟或，按一定的要求轉(zhuǎn)換成一個整數(shù)僚害。一定要把所有的要求都看清楚氏仗，不然test case 會有很多失敗的情況搀别。

從第一個非空格的字符開始算起碑隆，也就是前面的空格要trim掉
字符串可能含有任何字符猴贰，所以第一個非空字符可以是 +/- 或者數(shù)字麦轰，遇到其他的則停止
沒有有效的轉(zhuǎn)換淑倾，要輸出0
數(shù)字如果不處于 [?2^31, 2^31 ? 1] 之間馏鹤，則要輸出最大值或最小值，請參考例子5

個人感覺這道題的問題在于怎么樣包含所有可能的情況娇哆，有幾個重點的邊緣測試一定要考慮湃累，其他的問題應(yīng)該不是太難

Java Solution 1：

本人的辦法比較直接，沒有過多考慮時間碍讨，只是把所有能想到的case都cover住了治力。

public int myAtoi(String str) {
            final char[] acceptedArray = new char[]{'1','2','3','4','5','6','7','8','9','0'};
            final char[] acceptedSignedArray = new char[]{'1','2','3','4','5','6','7','8','9','0','+','-'};
            final int INT_MIN = Integer.MIN_VALUE;
            final int INT_MAX = Integer.MAX_VALUE;

            int num = 0;
            StringBuilder sb = new StringBuilder();

            //trim white space from str
            String trimedString = str.trim();

            if (trimedString == null || trimedString.length() == 0)
                return 0;

            char[] charArray = trimedString.toCharArray();

            //Retrieve the first char with Sign +/-
            char first = charArray[0];

            for(int i=0; i<acceptedSignedArray.length; i++ ){
                if (first == acceptedSignedArray[i]){
                    sb.append(first);
                    break;
                }
            }

            if(sb.length() == 0)
                return 0;

            for(int i = 1; i< charArray.length; i++){
                boolean bFound = false;
                for(int j=0; j<acceptedArray.length; j++ ){
                    if (charArray[i] == acceptedArray[j]){
                        sb.append(charArray[i]);
                        bFound = true;
                        break;
                    }
                }

                if(bFound == false)
                    break;

            }

        String s = sb.toString();
        //Filter sign only string
        if (s.equals("+") || s.equals("-"))
            return 0;

        //Filter out of int range string
        try {
            num = Integer.valueOf(s);
        }catch(NumberFormatException ex){
            if (s.indexOf("-") >= 0)
                return INT_MIN;

            return INT_MAX;
        }

        return num;
    }

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下面給出別人的解法

劃重點：

1.通過運算來生成整數(shù)，我是用string轉(zhuǎn)換
2.通過第一位是不是符號來決定是從 0勃黍，還是1 開始循環(huán)宵统，而我是單獨把第一位取出來

通過 1和-1來覺得符號的正負，這個想法不錯覆获。
判斷數(shù)字马澈，可以直接用大小判斷， chars[i] < '0' || chars[i] > '9'

public int myAtoi2(String str) {
        if (str == null) return 0;

        str = str.trim();
        if(str.length() == 0) return 0;

        char[] chars = str.toCharArray();

        // check if the 1st character is valid, i.e. must be a sign or a number
        if(chars[0] != '-' && chars[0] != '+' && chars[0] < '0' || chars[0] > '9') return 0;

        long sum = 0;
        int i = (chars[0] == '+' || chars[0] == '-') ? 1 : 0;
        int sign = chars[0] == '-' ? -1 : 1;
        for(; i < chars.length; i++){
            if(chars[i] < '0' || chars[i] > '9') break;
            else sum = sum * 10 + (chars[i] - '0');

            if(sum > Integer.MAX_VALUE && sign == 1)  return Integer.MAX_VALUE;
            if(sum > (long)Integer.MAX_VALUE + 1 && sign == -1) return Integer.MIN_VALUE;
        }

        return (int)sum * sign;

    }