Length of Last Word

https://leetcode.com/problems/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",
return 5.

解题思路：

从头算起，要考虑尾部的空格，比如"a "，还有尾部没有空格" a"，都要及时更新lastLength的数值。

public class Solution {
    public int lengthOfLastWord(String s) {
        int length = 0;
        int lastLength = 0;
        for(int i = 0; i < s.length(); i++){
            if(s.charAt(i) != ' '){
                length++;
            }else{
                if(length > 0){
                    lastLength = length;
                }
                length = 0;
            }
        }
        if(length > 0){
            lastLength = length;
        }
        return lastLength;
    }
}

后来想到还可以从后往前

public class Solution {
    public int lengthOfLastWord(String s) {
        int lastLength = 0;
        boolean metChar = false;
        for(int i = s.length() - 1; i >= 0; i--){
            if(s.charAt(i) == ' ' && !metChar){
                continue;
            }
            if(s.charAt(i) != ' '){
                lastLength++;
                metChar = true;
            }
            if(s.charAt(i) == ' ' && metChar){
                return lastLength;
            }
        }
        return lastLength;
    }
}

看到其他网友的解法，用双指针。仍然是从后往前遍历，lastWordIndex表示跳过结尾所有空格的第一个下标，firstSpaceIndex表示继续往前第一个单词前的空格下标。

public class Solution {
    public int lengthOfLastWord(String s) {
        int lastWordIndex = s.length() - 1;
        while(lastWordIndex >= 0 && s.charAt(lastWordIndex) == ' '){
            lastWordIndex--;
        }
        int firstSpaceIndex = lastWordIndex;
        while(firstSpaceIndex >= 0 && s.charAt(firstSpaceIndex) != ' '){
                firstSpaceIndex--;
        }
        return lastWordIndex - firstSpaceIndex;
    }
}