[LeetCode]--58. Length of Last Word

该博客讨论了如何解决LeetCode上的一个问题——找出字符串中最后一个单词的长度。作者提到,可以通过Java的库函数split实现,但如果不能使用库函数,则可以采用字符数组从后向前遍历的方法来检测非空字符,从而找到最后一个单词的长度。以'Hello World'为例,返回结果为5。

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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.

我利用java的库函数,split做的。

public int lengthOfLastWord(String s) {
        if (s == null || s.length() == 0)
            
### LeetCode Problem 58: Length of Last Word The goal is to find the length of the last word in a string. A word is defined as a maximal substring consisting of non-space characters only. #### Java Implementation Below is an efficient implementation using built-in methods: ```java class Solution { public int lengthOfLastWord(String s) { if (s == null || s.isEmpty()) return 0; String trimmedString = s.trim(); // Remove leading and trailing spaces[^3] if (trimmedString.isEmpty()) return 0; // Split by space, then get the last element's length. String[] words = trimmedString.split(" "); return words[words.length - 1].length(); } } ``` This code first checks if the input string `s` is either null or empty. If so, it returns zero immediately. Next, any leading and trailing whitespace from the string gets removed with `trim()`. Should this result be empty after trimming, again, zero is returned because no valid words exist. Finally, splitting the cleaned-up string into substrings based on spaces allows accessing the final array component which represents the last word whose length can thus be determined easily. For performance optimization considerations when dealing specifically with large strings where memory usage might become critical due to creating intermediate arrays during split operations, another approach directly iterates backward through the given string until encountering its initial non-whitespace character marking end-of-last-word boundary while counting letters encountered along the way without needing additional storage beyond single integer counter variable holding current count value.
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