LeetCode 65. Valid Number
7954 ワード
1.タイトルの説明
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.
2.問題の解き方
この問題は難しくないが,種々の制約関係,スペース,小数点,e,文字などの状況の処理を明らかにする必要がある.
3. code
4.大神解法
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.
2.問題の解き方
この問題は難しくないが,種々の制約関係,スペース,小数点,e,文字などの状況の処理を明らかにする必要がある.
3. code
class Solution {
public:
bool isNumber(string s) {
// handle the space
int pos = s.find_first_not_of(' ');
if (pos != -1)
s = s.substr(pos);
pos = s.find_last_not_of(' ');
if (-1 != pos)
s = s.substr(0, pos + 1);
bool hasNum = false, hasOper = false, hasE = false, hasPoint = false;
int id = 0;
int len_s = s.size();
while (id < len_s){
// +, - handle
if (id < len_s && s[id] == '+' || s[id] == '-'){
if (hasOper)
return false;
id++;
}
hasOper = true;
// num
if (id < len_s && s[id] >= '0' && s[id] <= '9'){
hasNum = true;
id++;
}
// has E, e
else if (id < len_s && (s[id] == 'e' || s[id] == 'E')){
if (!hasNum || hasE)
return false;
hasE = true;
hasOper = false;
hasNum = false;
id++;
}
// has point
else if (id < len_s && s[id] == '.'){
if (hasPoint || hasE)
return false;
id++;
hasPoint = true;
}
else{
return false;
}
}
return hasNum;
}
};
4.大神解法
/* All we need is to have a couple of flags so we can process the string in linear time: */
public boolean isNumber(String s) {
s = s.trim();
boolean pointSeen = false;
boolean eSeen = false;
boolean numberSeen = false;
boolean numberAfterE = true;
for(int i=0; i<s.length(); i++) {
if('0' <= s.charAt(i) && s.charAt(i) <= '9') {
numberSeen = true;
numberAfterE = true;
} else if(s.charAt(i) == '.') {
if(eSeen || pointSeen) {
return false;
}
pointSeen = true;
} else if(s.charAt(i) == 'e') {
if(eSeen || !numberSeen) {
return false;
}
numberAfterE = false;
eSeen = true;
} else if(s.charAt(i) == '-' || s.charAt(i) == '+') {
if(i != 0 && s.charAt(i-1) != 'e') {
return false;
}
} else {
return false;
}
}
return numberSeen && numberAfterE;
}
/* We start with trimming. If we see [0-9] we reset the number flags. We can only see . if we didn't see e or .. We can only see e if we didn't see e but we did see a number. We reset numberAfterE flag. We can only see + and - in the beginning and after an e any other character break the validation. At the and it is only valid if there was at least 1 number and if we did see an e then a number after it as well. So basically the number should match this regular expression: [-+]?[0-9]*(.[0-9]+)?(e[-+]?[0-9]+)? */