【Leetcode】Valid Number

題目：

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.

思路：

這題主要是在於輸入形式的多樣性，要進行周全的判斷，採用一些標識符是這題的解題的一大思路。

Java版本：

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;
}

思路是設置四個標識符 pointSeen = false，eSeen = false，numberSeen = false，numberAfterE = true，然後依次對字符串裏的字符進行判斷，並隨之更改標識符的狀態。

Another Java 版本：

public class Solution {
    public boolean isNumber(String s) {
        if (s == null) return false;
        
        s = s.trim();
        int n = s.length();
        
        if (n == 0) return false;
        
        // flags
        int signCount = 0;
        boolean hasE = false;
        boolean hasNum = false;
        boolean hasPoint = false;
        
        for (int i = 0; i < n; i++) {
            char c = s.charAt(i);
            
            // invalid character
            if (!isValid(c)) return false;
            
            // digit is always fine
            if (c >= '0' && c <= '9') hasNum = true;
            
            // e or E
            if (c == 'e' || c == 'E') {
                // e cannot appear twice and digits must be in front of it
                if (hasE || !hasNum) return false;
                // e cannot be the last one
                if (i == n - 1) return false;
                
                hasE = true;
            }
            
            // decimal place
            if (c == '.') {
                // . cannot appear twice and it cannot appear after e
                if (hasPoint || hasE) return false;
                // if . is the last one, digits must be in front of it, e.g. "7."
                if (i == n - 1 && !hasNum) return false;
                
                hasPoint = true;
            }
            
            // signs
            if (c == '+' || c == '-') {
                // no more than 2 signs
                if (signCount == 2) return false;
                // sign cannot be the last one
                if (i == n - 1) return false;
                // sign can appear in the middle only when e appears in front
                if (i > 0 && !hasE) return false;
                
                signCount++;
            }
        }
        
        return true;
    }
    
    boolean isValid(char c) {
        return c == '.' || c == '+' || c == '-' || c == 'e' || c == 'E' || c >= '0' && c <= '9';
    }
}

Python版本：

class Solution(object):
  def isNumber(self, s):
      """
      :type s: str
      :rtype: bool
      """
      #define a DFA
      state = [{}, 
              {'blank': 1, 'sign': 2, 'digit':3, '.':4}, 
              {'digit':3, '.':4},
              {'digit':3, '.':5, 'e':6, 'blank':9},
              {'digit':5},
              {'digit':5, 'e':6, 'blank':9},
              {'sign':7, 'digit':8},
              {'digit':8},
              {'digit':8, 'blank':9},
              {'blank':9}]
      currentState = 1
      for c in s:
          if c >= '0' and c <= '9':
              c = 'digit'
          if c == ' ':
              c = 'blank'
          if c in ['+', '-']:
              c = 'sign'
          if c not in state[currentState].keys():
              return False
          currentState = state[currentState][c]
      if currentState not in [3,5,8,9]:
          return False
      return True

這裏我們構造一個DFA（確定性有限狀態機），把字符串中各個組成的元素全提出到DFA中，然後，直接把s中的字符一一進行提取，並在DFA中進行判斷歸類。

C++版本：

class Solution {
public:
    bool isNumber(string str) {
        int state=0, flag=0; // flag to judge the special case "."
        while(str[0]==' ')  str.erase(0,1);//delete the  prefix whitespace 
        while(str[str.length()-1]==' ') str.erase(str.length()-1, 1);//delete the suffix whitespace
        for(int i=0; i<str.length(); i++){
            if('0'<=str[i] && str[i]<='9'){
                flag=1;
                if(state<=2) state=2;
                else state=(state<=5)?5:7;
            }
            else if('+'==str[i] || '-'==str[i]){
                if(state==0 || state==3) state++;
                else return false;
            }
            else if('.'==str[i]){
                if(state<=2) state=6;
                else return false;
            }
            else if('e'==str[i]){
                if(flag&&(state==2 || state==6 || state==7)) state=3;
                else return false;
            }
            else return false;
        }
        return (state==2 || state==5 || (flag&&state==6) || state==7);
    }
};

C++這個版本採用到的也是用到DFA這思路。輸入類型有5種，‘+,-’  ，數字 ， ‘.’，e，其他。另外採用到了state這個標識符有着8種狀態，是本思路的一大亮點。