【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",
return 5.

这道题让找最后一个单词的长度，所以我们采用双指针大发，第一个指针寻找最后一个单词出现的最后一个字母，第二个指针寻找最后一个单词出现的第一个字母的前面，两个指针一减就得到长度值了。

public int lengthOfLastWord(String s) {if (s == null || s.length() == 0)return 0;int index1 = s.length() - 1;while (index1 >= 0 && s.charAt(index1) == ' ')index1--;int index2 = index1;while (index2 >= 0 && s.charAt(index2) != ' ')index2--;return index1 - index2;}

如果这道题是让找第一个单词的长度呢？

大家可以思考一下，思路几乎一样

public static int lengthOfFirstWord(String s) {if (s == null || s.length() == 0)return 0;int index1 = 0;while (index1 < s.length() && s.charAt(index1) == ' ')index1++;int index2 = index1;while (index2 < s.length() && s.charAt(index2) != ' ')index2++;return index2 - index1;}