String to Integer (atoi) leetcode java

题目：

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.

题解：

这题主要就是考虑一下corner case。

越界问题？

正负号问题？

空格问题？

精度问题？

代码如下：

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 }

更新：

因为遇见过了atol，string to long 这样的问题，就不能用这种比当前数据类型长的方法解决。

另一种方法是更加普遍和巧妙的，就是用这样的判断：

1. Integer.MAX_VALUE/10 < result; //当前转换结果比Integer中最大值/10还大（因为这个判断放在while循环最开始，之后还要对result进行*10+当前遍历元素的操作，所以如果还乘10的result已经比Integer.MAX_VALUE/10还大，可想而知，乘了10更大）

2. Integer.MAX_VALUE/10 == result && Integer.MAX_VALUE%10 <(str.charAt(i) - '0') //另外的情况就是，当前result恰跟 Integer.MAX_VALUE/10相等，那么就判断当前遍历的元素值跟最大值的最后一位的大小关系即可

代码如下：

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 }

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

String to Integer (atoi) leetcode java相关推荐

008 String to Integer (atoi) [Leetcode]
题目内容: Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input ...
LeetCode算法入门- String to Integer (atoi)-day7
LeetCode算法入门- String to Integer (atoi)-day7 String to Integer (atoi): Implement atoi which converts ...
【细节实现题】LeetCode 8. String to Integer (atoi)
LeetCode 8. String to Integer (atoi) Solution1:我的答案参考链接:http://www.cnblogs.com/grandyang/p/4125537. ...
Kotlin实现LeetCode算法题之String to Integer (atoi)
题目String to Integer (atoi)(难度Medium) 大意是找出给定字串开头部分的整型数值,忽略开头的空格,注意符号,对超出Integer的数做取边界值处理. 方案1 1 clas ...
LeetCode 8. String to Integer (atoi)(字符串)
LeetCode 8. String to Integer (atoi)(字符串) LeetCode 8 String to Integer atoi字符串问题描述解题思路参考代码 By Sca ...
[LeetCode] NO. 8 String to Integer (atoi)
[题目] Implement atoi to convert a string to an integer. [题目解析] 该题目比较常见,从LeetCode上看代码通过率却只有13.7%,于是编码提 ...
leetcode 8. String to Integer (atoi)
也许是我没有理解清楚题意,为什么输入+-2的时候要输出0,而不是输出2呢. public class Solution {public int myAtoi(String str) {if(str = ...
8. String to Integer (atoi)
题目: Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input ca ...
Leet Code OJ 8. String to Integer (atoi) [Difficulty: Easy]
题目: Implement atoi to convert a string to an integer. Hint: Carefully consider all possible input ca ...

String to Integer (atoi) leetcode java

String to Integer (atoi) leetcode java相关推荐

最新文章

热门文章