LeetCode - 65. Valid Number
2016-07-02 21:04
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这道题目的条件描述非常不清晰,要求就是找出全部可能为数字的情况。由于自己太渣....这道题看了一会儿就放弃了,LeetCode Discuss上面有人的解法非常好,这里叙述一下他的思想,应该是基于正则表达式的。
首先我们把给定的字符串前后的空格都去掉,因为无论前后有多少空格,只要它的主体部分是数字,我们也还是将它认为成数字。接下来进行一系列的判断:
代码如下:
public class Solution{
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;
}
}
知识点:
1. String.trim()的左右是新建一个字符串,这个新的字符串的内容是将原来字符串开头和结尾的空格都去掉
首先我们把给定的字符串前后的空格都去掉,因为无论前后有多少空格,只要它的主体部分是数字,我们也还是将它认为成数字。接下来进行一系列的判断:
代码如下:
public class Solution{
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;
}
}
知识点:
1. String.trim()的左右是新建一个字符串,这个新的字符串的内容是将原来字符串开头和结尾的空格都去掉
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