leetcode-question65

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题目描述

验证给定的字符串是否为数字。
例如:
“0” => true
” 0.1 ” => true
“abc” => false
“1 a” => false
“2e10” => true
说明: 我们有意将问题陈述地比较模糊。在实现代码之前，你应当事先思考所有可能的情况。

思路：

主要是考虑各种可能的情况，如”+.” ” 123” 等等。

代码

class Solution {
    public boolean isNumber(String s) {
        boolean ret = false;
        if(s == null || s.equals("")){
            return ret;
        }
        if("+.".equals(s.trim())) return false;
        if("-.".equals(s.trim())) return false;
        char[] str2char = s.toCharArray();
        int start = 0;
        for(int i = 0; i < str2char.length;++i){
            if(str2char[i] == ' ') {
                start ++;
                continue;
            }
            break;
        }
        int end = str2char.length;
        for(int i = str2char.length -1 ; i >=0 ;-- i){
            if(str2char[i] == ' ') {
                end --;
                continue;
            }
            break;
        }
        if(start == str2char.length) return ret;
        int countE = 0;
        int countDot = 0;
        boolean hasE = false;
        boolean hasDot = false;
        int countP = 0;
        int countC = 0;
        int posE = 0;
        int posDot = 0;
        int posC = 0;
        int posP = 0;
        for(int i = start; i < end; ++ i){
            if(i == start && (str2char[i] > '9' || str2char[i] < '0') && str2char[i] != '-' && str2char[i] != '+' && str2char[i] != '.')
                return ret;
            if(i == start && (str2char[i] == '-' || str2char[i] == '+')){
                continue;
            }
            if((str2char[i] <= '9' && str2char[i] >= '0') || str2char[i] == 'e' || str2char[i] == 'E' || str2char[i] =='.' || 
               str2char[i] =='+' || str2char[i] == '-'){
                if(countE <= 1 && countDot <= 1 && str2char[i] <= '9' && str2char[i] >= '0'){
                    continue;
                }
                if(str2char[i] == 'e' || str2char[i] == 'E'){
                    hasE = true;
                    countE ++;
                    posE = i;
                    continue;
                }
                if(str2char[i] == '.'){
                    hasDot = true;
                    posDot = i;
                    countDot ++;
                    continue;
                }
                if(str2char[i] == '+'){
                    countP ++;
                    posP = i;
                    continue;
                }
                if(str2char[i] == '-'){
                    countC ++;
                    posC = i;
                    continue;
                }
            }
            if(str2char[i] > '9' || str2char[i] < '0' ){
                return ret;
            }   
        }
        if(countE > 1)
            return ret;
        if(countDot > 1)
            return ret;

        if(countP > 1){
            return ret;
        }
        if(countC > 1){
            return ret;
        }
        if((countP == 1 || countC == 1) && !hasE ) return ret;

        if(countP == 1 && posP == end - 1) return ret;
        if(countC == 1 && posC == end - 1) return ret;

        if(countP == 1 && hasE && posE == posP - 1 && posP != end - 1) return true;
        if(countC == 1 && hasE && posE == posC - 1 && posC != end - 1) return true;

        if(hasE && hasDot && posE < posDot) return ret;

        if(hasE && (posE == end - 1 || str2char[posE + 1] > '9' || str2char[posE + 1] < '0' || 
                    (hasDot && posDot==start && posE == posDot + 1) || 
                    (posE-1 < start || str2char[posE - 1] == '+' || str2char[posE - 1] =='-') )) 
            return ret;
        if(hasDot && ((posDot -1) >= start && 
                      ((str2char[posDot - 1] > '9' || str2char[posDot - 1] < '0') && 
                       (str2char[posDot - 1] != '+' && str2char[posDot - 1] != '-') ))) return ret;
        if(end - start == 1 && hasDot) return ret;

        return true;
    }
}