58. Length of Last Word
2016-09-29 22:07
246 查看
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
public class Solution {
public int lengthOfLastWord(String s) {
if(s=="")
return 0;
int len=s.length();
int i=len-1;
while(i>=0&&s.charAt(i)==' '){
--i;
}
if(i<0)
return 0;
int n=0;
while(i>=0&&s.charAt(i)!=' '){
++n;
--i;
}
return n;
}
}
WONDERFUL SOLUTION
public int lengthOfLastWord(String s) {
s = s.trim();
int lastIndex = s.lastIndexOf(' ') + 1;
return s.length() - lastIndex;
}
' ', 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.
public class Solution {
public int lengthOfLastWord(String s) {
if(s=="")
return 0;
int len=s.length();
int i=len-1;
while(i>=0&&s.charAt(i)==' '){
--i;
}
if(i<0)
return 0;
int n=0;
while(i>=0&&s.charAt(i)!=' '){
++n;
--i;
}
return n;
}
}
WONDERFUL SOLUTION
public int lengthOfLastWord(String s) {
s = s.trim();
int lastIndex = s.lastIndexOf(' ') + 1;
return s.length() - lastIndex;
}
相关文章推荐
- 58. Length of Last Word
- leetcode 58. Length of Last Word C++的stringstream的一个很简单的示例
- 58. Length of Last Word
- 【LeetCode】58. Length of Last Word
- leetcode: 58. Length of Last Word
- Leetcode刷题记——58. Length of Last Word(最后一个单词的长度)
- 58. Length of Last Word
- 58. Length of Last Word
- 【LeetCode】58. Length of Last Word
- 58. Length of Last Word
- leetcode58. Length of Last Word
- Easy-题目45:58. Length of Last Word
- [LeetCode] 58. Length of Last Word
- LeetCode 58. Length of Last Word
- Leetcode-58. Length of Last Word
- leetcode 58. Length of Last Word
- [Leetcode] 58. Length of Last Word 解题报告
- leetcode 58. Length of Last Word(java实现)
- [leetcode] 58. Length of Last Word 解题报告
- 58. Length of Last Word