leetcode 21: Length of Last Word

最新推荐文章于 2021-05-07 16:01:21 发布

原创最新推荐文章于 2021-05-07 16:01:21 发布 · 1.1k 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

216 篇文章

订阅专栏

本文将介绍如何通过编程解决字符串问题，具体为计算给定字符串中最后一个单词的长度。通过实例演示，理解如何处理字符串中的空格和边界情况。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

tips: 1. check corner situation.

2. if initial situation is different from others, deal with it before go into loop.

3. if ending situation is different from other, insert a sentinel or a dummy value.

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if( *s == '\0') return 0;
        
        int len = *s == ' ' ? 0 : 1;
        
        while(  (*++s) != '\0' ) {
            if( *s != ' ') {
                if( *(s-1) != ' ') {
                    len++;
                } else {
                    len = 1;
                }
            } 
        }
        
        return len;
    }
};

public class Solution {
    public int lengthOfLastWord(String s) {
        // Start typing your Java solution below
        // DO NOT write main() function
        // "  hello world    "
        
        if(s==null || s.length() <1) return 0;
        
        int count = s.charAt(0)==' ' ? 0 : 1;
        
        for(int i=1; i<s.length(); i++) {
            if(s.charAt(i) != ' ' ) {
                if(s.charAt(i-1) == ' ') {
                    count = 1;
                } else {
                    ++count;
                }
            }
        }
        
        return count;
    }
}