題目：

Implement atoi to convert a string to an integer.

Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.

Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.

spoilers alert... click to show requirements for atoi.

Requirements for atoi:

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. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.

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題解：

?這題主要就是考慮一下corner case。

?越界問題？

?正負號問題？

?空格問題？

?精度問題？

?代碼如下：

?

?1????public?int?atoi(String?str)?{
?2?????if?(str?==?null?||?str.length()?<?1)
?3?????????return?0;
?4??
?5?????//?trim?white?spaces
?6?????str?=?str.trim();
?7??
?8?????char?flag?=?'+';
?9??
10?????//?check?negative?or?positive
11?????int?i?=?0;
12?????if?(str.charAt(0)?==?'-')?{
13?????????flag?=?'-';
14?????????i++;
15?????}?else?if?(str.charAt(0)?==?'+')?{
16?????????i++;
17?????}
18?????//?use?double?to?store?result
19?????double?result?=?0;
20??
21?????//?calculate?value
22?????while?(str.length()?>?i?&&?str.charAt(i)?>=?'0'?&&?str.charAt(i)?<=?'9')?{
23?????????result?=?result?*?10?+?(str.charAt(i)?-?'0');
24?????????i++;
25?????}
26??
27?????if?(flag?==?'-')
28?????????result?=?-result;
29??
30?????//?handle?max?and?min
31?????if?(result?>?Integer.MAX_VALUE)
32?????????return?Integer.MAX_VALUE;
33??
34?????if?(result?<?Integer.MIN_VALUE)
35?????????return?Integer.MIN_VALUE;
36??
37?????return?(int)?result;
38?}

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更新：

因為遇見過了atol，string to long 這樣的問題，就不能用這種比當前數(shù)據(jù)類型長的方法解決。

另一種方法是更加普遍和巧妙的，就是用這樣的判斷：

1. Integer.MAX_VALUE/10 < result; //當前轉(zhuǎn)換結(jié)果比Integer中最大值/10還大（因為這個判斷放在while循環(huán)最開始，之后還要對result進行*10+當前遍歷元素的操作，所以如果還乘10的result已經(jīng)比Integer.MAX_VALUE/10還大，可想而知，乘了10更大）

2. Integer.MAX_VALUE/10 == result && Integer.MAX_VALUE%10 <(str.charAt(i) - '0') //另外的情況就是，當前result恰跟 Integer.MAX_VALUE/10相等，那么就判斷當前遍歷的元素值跟最大值的最后一位的大小關(guān)系即可

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代碼如下：

?1???public?int?atoi(String?str)?{
?2?????if?(str?==?null?||?str.length()?<?1)
?3?????????return?0;
?4??
?5?????//?trim?white?spaces?at?beginning?and?end
?6?????str?=?str.trim();
?7??
?8?????char?flag?=?'+';
?9??
10?????//?check?negative?or?positive
11?????int?i?=?0;
12?????if?(str.charAt(0)?==?'-')?{
13?????????flag?=?'-';
14?????????i++;
15?????}?else?if?(str.charAt(0)?==?'+')?{
16?????????i++;
17?????}
18?????//?use?double?to?store?result
19?????int?result?=?0;
20??
21?????//?calculate?value
22?????while?(str.length()?>?i?&&?str.charAt(i)?>=?'0'?&&?str.charAt(i)?<=?'9')?{
23?????????if(Integer.MAX_VALUE/10?<?result?||?(Integer.MAX_VALUE/10?==?result?&&?Integer.MAX_VALUE%10?<?(str.charAt(i)?-?'0')))?
24?????????????return?flag?==?'-'???Integer.MIN_VALUE?:?Integer.MAX_VALUE;
25?????????????
26?????????result?=?result?*?10?+?(str.charAt(i)?-?'0');
27?????????i++;
28?????}
29??
30?????if?(flag?==?'-')
31?????????result?=?-result;
32??
33?????return?result;
34?}

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Reference： http://www.programcreek.com/2012/12/leetcode-string-to-integer-atoi/

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