leetcode 58. 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.

Example:

Input: “Hello World”
Output: 5

给出一个string，要求return最后一个单词的长度，分隔符是space，如果没有最后一个单词，就返回0

乍一看很简单，不就是split

string[] result = s.split(" ");
return result[result.length - 1].length();

但是s=space时，result.length就会=0，然后就会ArrayIndexOutOfBounds

再者这种方法time上只超过60%

然后参照别人的一种方法，就是利用lastIndexOf(’ ')
并利用s.trim(),

但是注意s.trim()一定要保存到s，否则s是不变的
还有要注意是 s.length() -1 - index 而不是 s.length() - index，因为不包括‘ ’本身

    public int lengthOfLastWord(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        
        s = s.trim();
        
        int index = s.lastIndexOf(' ');
        
        return index == -1 ? s.length() : s.length() -1 - index;
    }