[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;
    }
}