LeetCode算法系列:65. Valid Number
阿新 • • 發佈:2018-12-10
目錄
題目描述:
Validate if a given string can be interpreted as a decimal number.
Some examples:
"0" => true " 0.1 " => true "abc" => false "1 a" => false "2e10" => true " -90e3 " => true " 1e" => false "e3" => false " 6e-1" => true " 99e2.5 " => false "53.5e93" => true " --6 " => false "-+3" => false "95a54e53" => false
Note: It is intended for the problem statement to be ambiguous. You should gather all requirements up front before implementing one. However, here is a list of characters that can be in a valid decimal number:
- Numbers 0-9
- Exponent - "e"
- Positive/negative sign - "+"/"-"
- Decimal point - "."
Of course, the context of these characters also matters in the input.
Update (2015-02-10):
The signature of the C++
function had been updated. If you still see your function signature accepts a const char *
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演算法實現:
總體來說比較簡單,只需要將正確的幾種形式弄清即可
(+/-)number(.number)(e number)
需要注意的是有一些情況需要根據除錯結果進行調整,比如空格的處理,比如,"1.",".2"都被認為是對的等
具體演算法中,博主維護了一個i來掃描整個字串,如下
class Solution {
public:
//跳過空格
void jumpblack(string s, int &i){
while(s[i] == ' ')i ++;
}
//跳過數字序列,如果有數字序列返回true,否則返回false
bool jumpdigits(string s, int &i){
int j = i;
while(s[i] != '\0' && s[i] >= '0' && s[i] <= '9')
i ++;
if(i > j)return true;
else return false;
}
//判斷e後面的序列合不合乎要求
bool judge_after_E(string s, int &i){
if(s[i] == '\0')return false;
if(s[i] == '+' || s[i] == '-') i ++;
if(s[i] == '\0')return false;
bool flag = jumpdigits(s, i);
jumpblack(s, i);
return s[i] == '\0' && flag;
}
bool isNumber(string s) {
if(s.empty())return false;
int i = 0;
jumpblack(s, i);
if(s[i] == '+' || s[i] == '-') i ++;
if(s[i] == '\0')return false;
bool flag = jumpdigits(s, i);
if(s[i] != '\0'){
if(s[i] == ' '){
jumpblack(s, i);
if(s[i] == '\0')return flag;
else return false;
}
else if(s[i] == '.'){
i ++;
bool flag1 = jumpdigits(s, i);
if(s[i] == ' ' || s[i] == '\0'){
jumpblack(s, i);
if(s[i] == '\0')return (flag || flag1);
else return false;
}
else if((flag1||flag) && s[i] == 'e')return judge_after_E(s, ++i);
else return false;
}
else if(flag && s[i] == 'e')return judge_after_E(s, ++ i);
else return false;
}
else return true;
}
};