[leetcode] Length of Last Word

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",
return5.

https://oj.leetcode.com/problems/length-of-last-word/

思路：从后向前扫描，定位lastword然后记录其长度。

public class Solution {
public int lengthOfLastWord(String s) {
if (s == null || s.length() < 1)
return 0;
int n = s.length();
int i;
int len = 0;
boolean in = false;
for (i = n - 1; i >= 0; i--) {
if (!in && s.charAt(i) != ' ') {
in = true;
}
if (in && s.charAt(i) != ' ') {
len++;
}
if (in && s.charAt(i) == ' ')
break;
}

return len;

}

public static void main(String[] args) {
System.out.println(new Solution().lengthOfLastWord("Hello World"));
System.out.println(new Solution().lengthOfLastWord("Hello  World  "));
System.out.println(new Solution().lengthOfLastWord("Hello  a  "));
System.out.println(new Solution().lengthOfLastWord(""));
System.out.println(new Solution().lengthOfLastWord("  "));

}
}

第二遍记录：

　　改了下代码，注意从后遍历时j>=0的越界判断不要忘记了。

public class Solution {
public int lengthOfLastWord(String s) {
if(s==null||s.length()==0)
return 0;
int n = s.length();
int j = n-1;
while(j>=0&&s.charAt(j)==' ')
j--;

int len =0;
while(j>=0&&s.charAt(j)!=' '){
len++;
j--;
}
return len;
}
}