資訊專欄INFORMATION COLUMN
摘要：兩種情況有無(wú)，如上所示。如何判斷是否只有一個(gè)之后的長(zhǎng)度小于等于。注意，為空只有在字符串最末位或者只有都是錯(cuò)誤的。規(guī)則若字符串首位為，取其，然后將應(yīng)用轉(zhuǎn)換為數(shù)組，逐位判斷字符，出現(xiàn)將置如已經(jīng)置，再次出現(xiàn)則直接返回。
Problem
Validate if a given string is numeric.
Example"0" => true " 0.1 " => true "abc" => false "1 a" => false "2e10" => trueNote
if !e A can be "+/-" + int + "." + int if e: AeB A can be "+/-" + int + "." + int B can be "+/-" + int
兩種情況：有無(wú)e，如上所示。
然后按有無(wú)e來(lái)拆分字符串，若拆分之后只有一部分，即用helper()函數(shù)判斷是否滿足構(gòu)成數(shù)的規(guī)則；若有兩部分，則對(duì)前一部分按數(shù)A（可以有小數(shù)部分）分析，后一部分按構(gòu)成數(shù)B分析（只能為整數(shù)）。
如何判斷是否只有一個(gè)e？split()之后的String[]長(zhǎng)度小于等于2。
由于數(shù)A和數(shù)B的區(qū)別只存在于有沒(méi)有小數(shù)點(diǎn)"."，建立一個(gè)flag hasDot()，其他應(yīng)用一樣的規(guī)則即可。注意，s為空、只有"e"、"e"在字符串最末位、或者只有"."都是錯(cuò)誤的。
規(guī)則：若字符串首位為+/-，取其.substring(1)，然后將string應(yīng)用toCharArray()轉(zhuǎn)換為char數(shù)組，逐位判斷字符，出現(xiàn)"."將hasDot()置true；如hasDot()已經(jīng)置true，再次出現(xiàn)"."則直接返回false。然后判斷其余位是否為數(shù)字即可。
Solutionpublic 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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