LeetCode -- Length of Last Word

Question:

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:

给出一个由大小写字母和空格' '构成的字符串,返回最后一个词的长度。

如果最后一个单词不存在,则返回0.

注意:判断是否是一个单词仅仅依靠是否有空格。

思路:首先将字符串按照空格划分到一个字符数组中,然后返回数组最后一个元素的长度即可。注意特殊情况:只有空格或者字符串为空。

 

Answer:

public class Solution {
    public int lengthOfLastWord(String s) {
        if(s == null)
            return 0;
        String [] str = s.split(" ");
        if(str.length == 0)
            return 0;
        String stri = str[str.length-1];
        char[] ch = stri.toCharArray();
        return ch.length;
    }
}

 

转载于:https://www.cnblogs.com/little-YTMM/p/4859175.html

### 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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