leetcode记录——65. 有效数字

有效数字（按顺序）可以分成以下几个部分：

1. 一个 小数 或者 整数
2. （可选）一个 'e' 或 'E' ，后面跟着一个 整数

小数（按顺序）可以分成以下几个部分：

1. （可选）一个符号字符（'+' 或 '-'）
2. 下述格式之一：
   1. 至少一位数字，后面跟着一个点 '.'
   2. 至少一位数字，后面跟着一个点 '.' ，后面再跟着至少一位数字
   3. 一个点 '.' ，后面跟着至少一位数字

整数（按顺序）可以分成以下几个部分：

1. （可选）一个符号字符（'+' 或 '-'）
2. 至少一位数字

部分有效数字列举如下：

• ["2", "0089", "-0.1", "+3.14", "4.", "-.9", "2e10", "-90E3", "3e+7", "+6e-1", "53.5e93", "-123.456e789"]

部分无效数字列举如下：

• ["abc", "1a", "1e", "e3", "99e2.5", "--6", "-+3", "95a54e53"]

给你一个字符串 s ，如果 s 是一个 有效数字 ，请返回 true 。

示例 1：

输入：s = "0"
输出：true

示例 2：

输入：s = "e"
输出：false

示例 3：

输入：s = "."
输出：false

示例 4：

输入：s = ".1"
输出：true

提示：

• 1 <= s.length <= 20
• s 仅含英文字母（大写和小写），数字（0-9），加号 '+' ，减号 '-' ，或者点 '.' 。

题解：

方法1：分类讨论

class Solution {

    public boolean isInteger(String s){

        if(s.length()==0)

            return false;

        int i=0;

        boolean ints=false;

        if(s.charAt(0)=='+'||s.charAt(0)=='-'){

            i+=1;

        }

        for(;i<s.length();++i){

            if(s.charAt(i)<'0'||s.charAt(i)>'9')

                return false;

            ints=true;

        }

        return ints;

    }

    public boolean isFloat(String s){

        if(s.length()==0)

            return false;

        boolean floats=false;

        boolean num=false;

        int i=0;

        if(s.charAt(0)=='+'||s.charAt(0)=='-'){

            i+=1;

        }

        for(;i<s.length();++i){

            if(s.charAt(i)=='.'&&floats==false){

                floats=true;

                continue;

            }

            if(s.charAt(i)<'0'||s.charAt(i)>'9')

                return false;

            num=true;

        }

        return num;

    }

    public boolean isNumber(String s) {

        if(s.length()==0)

            return false;

        String str=s.toLowerCase();

        String[] temp1 = str.split("e");

        if(temp1.length>=3||temp1.length<=0)

        {

            return false;

        }else{

            if(temp1.length==1){

                if(temp1[0].length()==s.length())

                    return (isFloat(temp1[0])||isInteger(temp1[0]));

                else

                    return false;

            }else{

                if(str.charAt(s.length()-1)=='e')

                return false;

                return (isFloat(temp1[0])||isInteger(temp1[0]))&&isInteger(temp1[1]);

            }

                

        }

    }

}

方法2：有限状态机

class Solution {
public boolean isNumber(String s) {
Map<State, Map<CharType, State>> transfer = new HashMap<State, Map<CharType, State>>();
Map<CharType, State> initialMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_INTEGER);
put(CharType.CHAR_POINT, State.STATE_POINT_WITHOUT_INT);
put(CharType.CHAR_SIGN, State.STATE_INT_SIGN);
}};
transfer.put(State.STATE_INITIAL, initialMap);
Map<CharType, State> intSignMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_INTEGER);
put(CharType.CHAR_POINT, State.STATE_POINT_WITHOUT_INT);
}};
transfer.put(State.STATE_INT_SIGN, intSignMap);
Map<CharType, State> integerMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_INTEGER);
put(CharType.CHAR_EXP, State.STATE_EXP);
put(CharType.CHAR_POINT, State.STATE_POINT);
}};
transfer.put(State.STATE_INTEGER, integerMap);
Map<CharType, State> pointMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_FRACTION);
put(CharType.CHAR_EXP, State.STATE_EXP);
}};
transfer.put(State.STATE_POINT, pointMap);
Map<CharType, State> pointWithoutIntMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_FRACTION);
}};
transfer.put(State.STATE_POINT_WITHOUT_INT, pointWithoutIntMap);
Map<CharType, State> fractionMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_FRACTION);
put(CharType.CHAR_EXP, State.STATE_EXP);
}};
transfer.put(State.STATE_FRACTION, fractionMap);
Map<CharType, State> expMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_EXP_NUMBER);
put(CharType.CHAR_SIGN, State.STATE_EXP_SIGN);
}};
transfer.put(State.STATE_EXP, expMap);
Map<CharType, State> expSignMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_EXP_NUMBER);
}};
transfer.put(State.STATE_EXP_SIGN, expSignMap);
Map<CharType, State> expNumberMap = new HashMap<CharType, State>() {{
put(CharType.CHAR_NUMBER, State.STATE_EXP_NUMBER);
}};
transfer.put(State.STATE_EXP_NUMBER, expNumberMap);

int length = s.length();
State state = State.STATE_INITIAL;

for (int i = 0; i < length; i++) {
CharType type = toCharType(s.charAt(i));
if (!transfer.get(state).containsKey(type)) {
return false;
} else {
state = transfer.get(state).get(type);
}
}
return state == State.STATE_INTEGER || state == State.STATE_POINT || state == State.STATE_FRACTION || state == State.STATE_EXP_NUMBER || state == State.STATE_END;
}

public CharType toCharType(char ch) {
if (ch >= '0' && ch <= '9') {
return CharType.CHAR_NUMBER;
} else if (ch == 'e' || ch == 'E') {
return CharType.CHAR_EXP;
} else if (ch == '.') {
return CharType.CHAR_POINT;
} else if (ch == '+' || ch == '-') {
return CharType.CHAR_SIGN;
} else {
return CharType.CHAR_ILLEGAL;
}
}

enum State {
STATE_INITIAL,
STATE_INT_SIGN,
STATE_INTEGER,
STATE_POINT,
STATE_POINT_WITHOUT_INT,
STATE_FRACTION,
STATE_EXP,
STATE_EXP_SIGN,
STATE_EXP_NUMBER,
STATE_END
}

enum CharType {
CHAR_NUMBER,
CHAR_EXP,
CHAR_POINT,
CHAR_SIGN,
CHAR_ILLEGAL
}
}

方法3：正则匹配

class Solution {
public:
    static const regex pattern;

    bool isNumber(string str) {
        return regex_match(str, pattern);
    }
};

const regex Solution::pattern("[+-]?(?:\d+\.?\d*|\.\d+)(?:[Ee][+-]?\d+)?");