[Leetcode]58. Length of Last Word

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。该方法通过C++实现,利用STL中的<algorithm>库函数find_if和find_if_not进行非空字符和字母的查找,并使用distance计算两查找结果间的距离。

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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.

class Solution {
public:
    int lengthOfLastWord(string s) {
        auto first = find_if(s.rbegin(), s.rend(), ::isalpha);
        auto last = find_if_not(first, s.rend(), ::isalpha);
        return distance(first, last);
    }
};
### 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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