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[?]*Valid Number

时间:2016-01-09 06:17:23      阅读:300      评论:0      收藏:0      [点我收藏+]

Validate if a given string is numeric.

Some examples:
"0" => true
" 0.1 " => true
"abc" => false
"1 a" => false
"2e10" => true

Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button  to reset your code definition.

 

/**
 * 本代码由九章算法编辑提供。没有版权欢迎转发。
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 */

// Non-regex version

public class Solution {
    public boolean isNumber(String s) {
        int len = s.length();
        int i = 0, e = len - 1;
        while (i <= e && Character.isWhitespace(s.charAt(i))) i++;
        if (i > len - 1) return false;
        while (e >= i && Character.isWhitespace(s.charAt(e))) e--;
        // skip leading +/-
        if (s.charAt(i) == ‘+‘ || s.charAt(i) == ‘-‘) i++;
        boolean num = false; // is a digit
        boolean dot = false; // is a ‘.‘
        boolean exp = false; // is a ‘e‘
        while (i <= e) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                num = true;
            }
            else if (c == ‘.‘) {
                if(exp || dot) return false;
                dot = true;
            }
            else if (c == ‘e‘) {
                if(exp || num == false) return false;
                exp = true;
                num = false;
            }
            else if (c == ‘+‘ || c == ‘-‘) {
                if (s.charAt(i - 1) != ‘e‘) return false;
            }
            else {
                return false;
            }
            i++;
        }
        return num;
    }
}

 

CleanCode Version: ?

[?]*Valid Number

原文:http://www.cnblogs.com/hygeia/p/5115387.html

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