LeetCode 58. Length of Last Word

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本文介绍了一个使用C++实现的简单算法，该算法用于计算给定字符串中最后一个单词的长度。通过对字符串从后向前遍历，忽略尾部空格并找到单词边界，能够有效地计算出最后一个单词的长度。

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.

Two pointers easy version.

#include <string>
#include <iostream>
using namespace std;

int lengthOfLastWord(string s) {
    if(s.size() == 0) return 0;
    int i = s.size() - 1;
    while(i >= 0 && s[i] == ' ') {
        i--;
    }
    int j = i;
    while(j >= 0 && s[j] != ' ') {
        j--;
    }
    return i - j;
}

int main(void) {
    int len = lengthOfLastWord("a dc ");
    cout << len << endl;
}