58. Length of Last Word

本文介绍了一种在Java中获取字符串最后一个单词长度的方法。通过使用trim()去除字符串两端的空白字符,再利用split()按空格分割字符串,从而准确获取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.


一看题目也太简单了吧!先按照空格split以下然后去最后一个字符串长度不就行了!!但是。。

居然出错了 说indexoutofbound

参考了别人的代码,在split之前加了一个trim去掉开头结尾的空格终于AC

原因分析:如果最后有多个空格,那么会出现最后字符串是空格的结果 。。。解释不通 再想想吧

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


### 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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