【LeetCode】#65有效数字(Valid Number)

【LeetCode】#65有效数字(Valid Number)

题目描述

验证给定的字符串是否为数字。
说明: 我们有意将问题陈述地比较模糊。在实现代码之前，你应当事先思考所有可能的情况。

示例

“0” => true
" 0.1 " => true
“abc” => false
“1 a” => false
“2e10” => true

Description

Validate if a given string can be interpreted as a decimal number.
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:
1.Numbers 0-9
2.Exponent - “e”
3.Positive/negative sign - “+”/"-"
4.Decimal point - “.”
Of course, the context of these characters also matters in the input.

Example

“0” => true
" 0.1 " => true
“abc” => false
“1 a” => false
“2e10” => true
" -90e3 " => true
" 1e" => false
“e3” => false
" 6e-1" => true
" 99e2.5 " => false
“53.5e93” => true
" --6 " => false
“-+3” => false
“95a54e53” => false

解法

class Solution {
    public boolean isNumber(String s) {
        s = s.trim();
        boolean pointSeen = false;
        boolean eSeen = false;
        boolean numberSeen = false;
        boolean numberAfterE = false;
        for(int i=0; i<s.length(); i++){
            if('0'<=s.charAt(i) && s.charAt(i)<='9'){
                numberSeen = true;
                numberAfterE = true;
            }else if(s.charAt(i)=='.'){
                if(eSeen || pointSeen){
                    return false;
                }
                pointSeen = true;
            }else if(s.charAt(i)=='e'){
                if(eSeen || !numberAfterE){
                    return false;
                }
                numberAfterE = false;
                eSeen = true;
            }else if(s.charAt(i)=='-' || s.charAt(i)=='+'){
                if(i!=0 && s.charAt(i-1)!='e'){
                    return false;
                }
            }else{
                return false;
            }
        }
        return numberSeen && numberAfterE;
    }
}