【Leetcode】Valid Number

题目：

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：

这题主要是在于输入形式的多样性，要进行周全的判断，采用一些标识符是这题的解题的一大思路。

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;}

思路是设置四个标识符 pointSeen = false，eSeen = false，numberSeen = false，numberAfterE = true，然后依次对字符串里的字符进行判断，并随之更改标识符的状态。

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

这里我们构造一个DFA（确定性有限状态机），把字符串中各个组成的元素全提出到DFA中，然后，直接把s中的字符一一进行提取，并在DFA中进行判断归类。

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++这个版本采用到的也是用到DFA这思路。输入类型有5种，‘+,-’  ，数字 ， ‘.’，e，其他。另外采用到了state这个标识符有着8种状态，是本思路的一大亮点。