Leetcode 65. Valid Number

题目

65. Valid Number

Difficulty: Hard
Total Accepted: 99.5K
Total Submissions: 753.7K

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.

解题报告

AC 截图

题目大意

检查一个字符串的内容（在删去前后缀的空格后）是否是一个合法的十进制数。

解题思路

该题可以通过将字符串分块来令问题化为简单的子问题的组合。首先考虑有 E 的字符串，则我们只需要判断左边是否为一个合法小数，右边是否为一个合法整数。

对于一个小数字符串，我们只需要判断小数点左边和右边是否为一个合法整数。

对于一个整数字符串，我们只需要判断除了符号外，其他字符是否是一个合法数字即可。

坑点

1. 和 .3 都是合法的数字。

题解

AC 截图

C++ 代码

class Solution {
 public:
  bool isNumber(string s) {
    int front = 0, last = s.size() - 1;
    for (; front < last && s[front] == ' '; ++front);
    for (; front < last && s[last] == ' '; --last);
    s = s.substr(front, last - front + 1);

    int indexOfE;
    if ((indexOfE = s.find('e')) != -1) {
      if (indexOfE != s.size() - 1) {
        return (isFraction(s.substr(0, indexOfE)) &&
          isInteger(s.substr(indexOfE + 1), true));
      }
      return false;
    } else {
      return isFraction(s);
    }
    return false;
  }

  bool isFraction(string s) {
    s = s.substr(s[0] == '+' || s[0] == '-');
    int indexOfDot = s.find('.');
    if (indexOfDot != -1 && indexOfDot != s.size() - 1) {
      return isInteger(s.substr(0, indexOfDot), false, true) && isInteger(s.substr(indexOfDot + 1), false, true);
    } 
    return isInteger(s);
  }

  bool isInteger(string s, bool checkMinus = false, bool allowZero = false) {
    if (checkMinus) {
      s = s.substr(s[0] == '+' || s[0] == '-');
    }

    for (const auto &iter : s) {
      if (iter > '9' || iter < '0') {
        return false;
      }
    }

    return allowZero || s.size();
  }
};