[LintCode] Valid Number

since1986 发布于2019-08-14 15:15 / 2464人阅读

摘要：两种情况有无，如上所示。如何判断是否只有一个之后的长度小于等于。注意，为空只有在字符串最末位或者只有都是错误的。规则若字符串首位为，取其，然后将应用转换为数组，逐位判断字符，出现将置如已经置，再次出现则直接返回。

Problem

Validate if a given string is numeric.

Example

"0" => true

" 0.1 " => true

"abc" => false

"1 a" => false

"2e10" => true

Note

if !e
A can be "+/-" + int + "." + int

if e: AeB
A can be "+/-" + int + "." + int
B can be "+/-" + int

两种情况：有无e，如上所示。
然后按有无e来拆分字符串，若拆分之后只有一部分，即用helper()函数判断是否满足构成数的规则；若有两部分，则对前一部分按数A（可以有小数部分）分析，后一部分按构成数B分析（只能为整数）。
如何判断是否只有一个e？split()之后的String[]长度小于等于2。
由于数A和数B的区别只存在于有没有小数点"."，建立一个flag hasDot()，其他应用一样的规则即可。注意，s为空、只有"e"、"e"在字符串最末位、或者只有"."都是错误的。

规则：若字符串首位为+/-，取其.substring(1)，然后将string应用toCharArray()转换为char数组，逐位判断字符，出现"."将hasDot()置true；如hasDot()已经置true，再次出现"."则直接返回false。然后判断其余位是否为数字即可。

Solution

public class Solution {

public boolean isNumber(String s) {
    s = s.trim();
    if (s == null || (s.length() > 0 && s.charAt(s.length()-1) == "e") || s.equals(".")) return false;
    String[] strs = s.split("e");
    if (strs.length > 2) return false;
    boolean res = helper(strs[0], false);
    if (strs.length == 2) {
        res = res && helper(strs[1], true);
    }
    return res;
}
public boolean helper(String s, boolean hasDot) {
    if (s.length() > 0 && (s.charAt(0) == "+" || s.charAt(0) == "-"))
        s = s.substring(1);
        char[] ch = s.toCharArray();
        if (ch.length == 0) return false;
        for (int i =  0; i < ch.length; i++) {
            if (ch[i] == ".") {
                if (hasDot) return false;
                hasDot = true;
            }
            else if (!(ch[i] >= "0" && ch[i] <= "9")) return false;
        }
        return true;
}

}

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