[LeetCode]65.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.

Update (2015-02-10):
The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.

思路

无

代码

/*---------------------------------------*   日期：2015-06-16*   作者：SJF0115*   题目: 65.Valid Number*   网址：https://leetcode.com/problems/valid-number/*   结果：AC*   来源：LeetCode*   博客：-----------------------------------------*/#include <iostream>#include <cstdio>using namespace std;class Solution {public:    bool isNumber(string s) {        int size = s.size();        if(size == 0){            return false;        }//if        // 前导0        int start = 0;        while(s[start] == ' '){            ++start;        }//while        // 后导0        int end = size - 1;        while(s[end] == ' '){            --end;        }//while        bool hasNum = false,hasPoint = false,hasE = false;        for(int i = start;i <= end;++i){            if(s[i] == '.'){                // 如果前面已经有了'.' 或者 'e'                if(hasPoint || hasE){                    return false;                }//if                hasPoint = true;            }//if            else if(s[i] == 'e'){                // 如果前面已经有了'e' 或者 没数字                if(hasE || !hasNum){                    return false;                }//if                hasE = true;            }//else            else if(s[i] < '0' || s[i] > '9'){                // +2                if(i == start && (s[i] == '+' || s[i] == '-')){                    continue;                }//if                // 1e-2                else if((i != 0 && s[i-1] == 'e') && (s[i] == '+' || s[i] == '-')){                    continue;                }                else{                    return false;                }//else            }//else            else{                hasNum = true;            }//else        }//for        // 最后有效位不能是'e+-'        if(s[end] == 'e' || s[end] == '+' || s[end] == '-'){            return false;        }//if        // '.'        if(!hasNum && hasPoint){            return false;        }//if        // 全是空格        if(end == -1){            return false;        }//if        return true;    }};int main(){    Solution s;    string str("    4.4e3   ");    //char* str = ".3e4";    cout<<s.isNumber(str)<<endl;    return 0;}

运行时间

测试用例

正确：
.3
3.
3.3
4e4
46.e3
4.3e3
.4e4

错误：
e
.
4e
e4
.e4
45e.2