• 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.

    这题还是比较容易的,用正则表达式感觉更简单一点,代码如下:

    import java.util.regex.*;
    public class Solution {
        public int lengthOfLastWord(String s) {
            ArrayList<String> word_list = new ArrayList<String>();
            Pattern p = Pattern.compile("\w+");//匹配正则表达式
            Matcher m = p.matcher(s);
            while(m.find()){
                word_list.add(m.group());
            }
            int size = word_list.size();
            //size==0表示没有找到任何单词
            if(size==0){
                return 0;
            }
            return word_list.get(size-1).length();
        }
    }

     不用正则表达式代码:

    public class Solution {
        public int lengthOfLastWord(String s) {
           if(s.length()==0){
               return 0;
           }
           char[] array = s.toCharArray();
           int size = array.length;
           ArrayList<String> list = new ArrayList<String>();
           String word = "";//临时存储一个单词
           int flag = 0;//flag==1时表示当前正在记录一个单词
           for(int i=0;i<size;i++){
               if(array[i]!=' '){
                  word = word+array[i];
                  flag = 1;//修改flag表示当前开始记录一个单词
               }
               //如果遇到空格或者循环结束则将flag置0,表示当前单词记录结束
               if(array[i]==' '||i==size-1){
                   flag = 0;
               }
               //如果word非空并且flag==0则表示一个单词记录完毕并将其存入list中,同时word置空
               if(!word.isEmpty()&&flag==0){
                   list.add(word);
                   word = "";
               }
           }
           if(list.isEmpty()){
               return 0;
           }
           return list.get(list.size()-1).length();
        }
    }
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  • 原文地址:https://www.cnblogs.com/mrpod2g/p/4269710.html
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