题目：

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.

题意：

难度不大，考虑所有边界情况即可。

解法：

C++版本实现方法1：

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   // 从右往左依次遍历即可
  class Solution {
	public:
		int lengthOfLastWord(string s) {
			int length = 0;
			for (int index = s.length() - 1; index >= 0; index--){
				char c = s.at(index);
				if (c != ' '){
					length++;
				}
				else if (c == ' ' && length != 0){
					return length;
				}
			}
			return length;
		}
	}

leetCode Oj系统评判结果如下图所示：

leetCode C++1;

C++版本实现方法2：

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	// 直接利用现有轮子STL 
class Solution {
	public:
		int lengthOfLastWord(string s) {
			auto left = find_if(s.rbegin(), s.rend(), ::isalpha);
			auto right = find_if_not(left, s.rend(), ::isalpha);
			return distance(left, right);
		}
	}

leetCode Oj系统评判结果如下图所示：

leetCode C++2;