Length of Last Word
2017-07-12 00:00
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问题:
Given a string s consists of upper/lower-case alphabets and empty space characters
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
For example,
Given s =
return
解决:
① 将字符串转换为字符串数组,然后计算。
public class Solution {//7ms
public int lengthOfLastWord(String s) {
String[] schar = s.trim().split(" ");
if(schar.length == 0) return 0;
return schar[schar.length - 1].length();
}
}
② 直接在字符串末尾开始计算,停在‘ ’的位置。
public class Solution {//5ms
public int lengthOfLastWord(String s) {
int len = 0;
for(int i = s.length() - 1; i >= 0; i --){
if(s.charAt(i) != ' ')
len ++;
else if(len != 0)
break;
}
return len;
}
}
Given a string s consists of upper/lower-case alphabets and empty space characters
' ', return the length of last word in the string.
If the last word does not exist, return 0.
Note: A word is defined as a character sequence consists of non-space characters only.
For example,
Given s =
"Hello World",
return
5.
解决:
① 将字符串转换为字符串数组,然后计算。
public class Solution {//7ms
public int lengthOfLastWord(String s) {
String[] schar = s.trim().split(" ");
if(schar.length == 0) return 0;
return schar[schar.length - 1].length();
}
}
② 直接在字符串末尾开始计算,停在‘ ’的位置。
public class Solution {//5ms
public int lengthOfLastWord(String s) {
int len = 0;
for(int i = s.length() - 1; i >= 0; i --){
if(s.charAt(i) != ' ')
len ++;
else if(len != 0)
break;
}
return len;
}
}
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