LeetCode 66 Valid Number

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1：使用有限状态机，空间复杂度为O(n),时间复杂度为O(n)，本思路来自龚陆安(http://weibo.com/luangong)，只不过原作者使用的是c++，我将其改写为java。

public class Solution {
	public boolean isNumber(String s) {
		int transitionTable[][] = new int[][] {
				{ -1, 0, 3, 1, 2, -1 }, // next states for state 0
				{ -1, 8, -1, 1, 4, 5 }, // next states for state 1
				{ -1, -1, -1, 4, -1, -1 }, // next states for state 2
				{ -1, -1, -1, 1, 2, -1 }, // next states for state 3
				{ -1, 8, -1, 4, -1, 5 }, // next states for state 4
				{ -1, -1, 6, 7, -1, -1 }, // next states for state 5
				{ -1, -1, -1, 7, -1, -1 }, // next states for state 6
				{ -1, 8, -1, 7, -1, -1 }, // next states for state 7
				{ -1, 8, -1, -1, -1, -1 } // next states for state 8
		};
		int state = 0;
		for (int i = 0; i < s.length(); ++i) {
			int inputType = 0;
			if (Character.isSpaceChar(s.charAt(i)))
				inputType = 1;
			else if (s.charAt(i) == '+' || s.charAt(i) == '-')
				inputType = 2;
			else if (Character.isDigit(s.charAt(i)))
				inputType = 3;
			else if (s.charAt(i) == '.')
				inputType = 4;
			else if (s.charAt(i) == 'e' || s.charAt(i) == 'E')
				inputType = 5;
			// Get next state from current state and input symbol
			state = transitionTable[state][inputType];
			// Invalid input
			if (state == -1)
				return false;
		}
		// If the current state belongs to one of the accepting (final) states,
		// then the number is valid
		return state == 1 || state == 4 || state == 7 || state == 8;
	}

}

思路2 java中可以使用正则表达式，我们可以使用正则表达式来处理。

public class Solution {
	public boolean isNumber(String s) {
	    //首先删除字符串首位和尾部的空格
	    String str=s.replaceAll("(^\\s+)|(\\s+$)", "");
	
		if(str.length()==0) return false;
		
		boolean [] temp=new boolean[6];
		boolean result=false;
		temp[0]=str.matches("(-|\\+)?\\d+(\\.)?");
		
		temp[1]=str.matches("(-|\\+)?\\d+\\.\\d+");
		temp[2]=str.matches("(-|\\+)?\\.\\d+");
		
		temp[3]=str.matches("(-|\\+)?\\d+(\\.)?(e|E)(-|\\+)?\\d+");
		temp[4]=str.matches("(-|\\+)?\\d+\\.\\d+(e|E)(-|\\+)?\\d+");
		temp[5]=str.matches("(-|\\+)?\\.\\d+(e|E)(-|\\+)?\\d+");
		
		for(int i=0;i<6;i++){
			result=result||temp[i];
		}
		return result;
	}
}

思路3：与思路2相同，只不过对正则表达式进行了精简。

public boolean isNumber(String s) {
		 String str=s.trim();//删除两端的空格
		if(str.length()==0) return false;
		
		boolean [] temp=new boolean[4];
		temp[0]=str.matches("(-|\\+)?\\d+(\\.)?");
		temp[1]=str.matches("(-|\\+)?(\\d+)?(\\.)?\\d+");	
		temp[2]=str.matches("(-|\\+)?\\d+(\\.)?(e|E)(-|\\+)?\\d+");
		temp[3]=str.matches("(-|\\+)?(\\d+)?(\\.)?\\d+(e|E)(-|\\+)?\\d+");	
		return temp[0]||temp[1]||temp[2]||temp[3];
	}