[leet code] Length of Last Word

最新推荐文章于 2025-08-20 17:55:57 发布

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最新推荐文章于 2025-08-20 17:55:57 发布

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分类专栏： interview preparation 文章标签： string algorithm java leetcode

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本文介绍了一种方法来计算给定字符串中最后一个单词的长度。该方法不使用内置的split函数，而是通过从字符串的末尾开始遍历并跳过空格，找到最后一个单词的起始位置，从而计算出最后一个单词的长度。

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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.

========

Analysis:

Utilize .split to find out the last word, then return the length of the last word.

public class Solution {
    public int lengthOfLastWord(String s) {
        String[] array = s.split(" ");
        if (array.length == 0) return 0;
        String last = array[array.length-1];
        return last.length();
    }
}

Dont utilize any build-in method, we can

1. Find out the first nun-space character from the right most element of the array. (deal with the case like "abc def ")

2. (Based on 1.) Length++ until the first space reached or 1st element of the array reached

public class Solution {
    public int lengthOfLastWord(String s) {
        if(s.length() == 0) return 0;
        
        int length = 0;
        int i=s.length()-1;
        while(i>=0&&s.charAt(i) == ' ') i--; // find out last character of last word
        while(i>=0&&s.charAt(i) != ' ') {
            i--;
            length++;
        }
        return length;
    }
}