题目

Validate if a given string can be interpreted as a decimal number.

Some examples:
"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

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:

  • Numbers 0-9
  • Exponent - “e”
  • Positive/negative sign - “+”/“-“
  • Decimal point - “.”

Of course, the context of these characters also matters in the input.

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.

题意

给定一个字符串，判断它能不能正确表示一个数。满足条件的字符串中只可能包含以下字符：’0’-‘9’、科学计数’e’、正负符号’+’/‘-‘、小数点’.’。

思路

坑点在于需要判断的情况比较多。我的方法是针对字符串中的每一个字符，只判断它与前后两个字符形成的组合是否有效。具体有效组合参考注释。(d代表数字)

代码实现

Java

  1. class Solution {
  2.     public boolean isNumber(String s) {
  3.         boolean isExponentExist = false;    // 记录'e'是否已出现
  4.         boolean isPointExist = false;        // 记录'.'是否已出现
  6.         s = s.trim();
  8.         if (s.length() == 0) {
  9.             return false;
  10.         }
  12.         for (int i = 0; i < s.length(); i++) {
  13.             char c = s.charAt(i);
  14.             if (!isDigit(c) && c != '+' && c != '-' && c != 'e' && c != '.') {
  15.                 return false;
  16.             }
  17.             if ((c == '+' || c == '-') && !isPlusOrMinusValid(s, i)) {
  18.                 return false;
  19.             }
  20.             if (c == 'e') {
  21.                 if (isExponentExist || !isExponentValid(s, i)) {
  22.                     return false;
  23.                 }
  24.                 isExponentExist = true;
  25.             }
  26.             if (c == '.') {
  27.                 if (isPointExist || isExponentExist || !isPointValid(s, i)) {
  28.                     return false;
  29.                 }
  30.                 isPointExist = true;
  31.             }
  32.         }
  33.         return true;
  34.     }
  36.     private boolean isDigit(char c) {
  37.         return c >= '0' && c <= '9';
  38.     }
  40.     // '+'/'-'的有效组合:   [+-][d.]  e[+-]d
  41.     private boolean isPlusOrMinusValid(String s, int i) {
  42.         if (i == 0 && 1 < s.length() && (isDigit(s.charAt(1)) || s.charAt(1) == '.')) {
  43.             return true;
  44.         }
  45.         if (i - 1 >= 0 && s.charAt(i - 1) == 'e' && i + 1 < s.length() && isDigit(s.charAt(i + 1))) {
  46.             return true;
  47.         }
  48.         return false;
  49.     }
  51.     // 'E'的有效组合:   [d.]e[d+-] 
  52.     private boolean isExponentValid(String s, int i) {
  53.         if (i - 1 >= 0) {
  54.             char left = s.charAt(i - 1);
  55.             if (i + 1 < s.length()) {
  56.                 char right = s.charAt(i + 1);
  57.                 if ((isDigit(left) || left == '.') && (isDigit(right) || right == '+' || right == '-')) {
  58.                     return true;
  59.                 }
  60.             }
  61.         }
  62.         return false;
  63.     }
  65.     // '.'的有效组合:   d.d  d.e  [+-].d  d.  .d     
  66.     private boolean isPointValid(String s, int i) {
  67.         if (i - 1 >= 0) {
  68.             char left = s.charAt(i - 1);
  69.             if (i + 1 < s.length()) {
  70.                 char right = s.charAt(i + 1);
  71.                 if (isDigit(left) && (right == 'e' || isDigit(right))) {
  72.                     return true;
  73.                 }
  74.                 if ((left == '+' || left == '-') && isDigit(right)) {
  75.                     return true;
  76.                 }
  77.             } else {
  78.                 if (isDigit(left)) {
  79.                     return true;
  80.                 }
  81.             }
  82.         } else {
  83.             if (i + 1 < s.length()) {
  84.                 char right = s.charAt(i + 1);
  85.                 if (isDigit(right)) {
  86.                     return true;
  87.                 }
  88.             }
  89.         }
  90.         return false;
  91.     }
  92. }

JavaScript

  1. /**
  2.  * @param {string} s
  3.  * @return {boolean}
  4.  */
  5. var isNumber = function(s) {
  6.     return !s.match(/.?Infinity/) && !Number.isNaN(Number(s))
  7. };