Valid Number Java

最新推荐文章于 2021-06-21 17:15:44 发布

转载最新推荐文章于 2021-06-21 17:15:44 发布 · 340 阅读

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本文介绍了一种判断给定字符串是否为有效数值的方法，并提供了一个清晰的Java实现案例，详细解析了如何处理各种边界情况。

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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.

作为一道目前AP绿最低的一道题，着实蛋疼，感觉就是需要考虑的情况比较多，自己尝试多次之后毅然去了讨论区，找了一份如下代码：

源码地址：https://leetcode.com/discuss/26682/clear-java-solution-with-ifs

public boolean isNumber(String s) {
    s = s.trim();

    boolean pointSeen = false;
    boolean eSeen = false;
    boolean numberSeen = false;
    boolean numberAfterE = true;
    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 || !numberSeen) {
                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;
}

并且附上原文解释：
We start with trimming.

If we see [0-9] we reset the number flags.
We can only see . if we didn’t see e or ..
We can only see e if we didn’t see e but we did see a number. We reset numberAfterE flag.
We can only see + and - in the beginning and after an e
any other character break the validation.
At the and it is only valid if there was at least 1 number and if we did see an e then a number after it as well.

So basically the number should match this regular expression:

[-+]?[0-9]*(.[0-9]+)?(e[-+]?[0-9]+)?