LeetCode 58:Length of Last Word

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.

方法1：正序遍历整个字符串，碰到非空格字符‘ ’ ，len++，碰到空格字符且后一个字符为非空格字符，则len清零

class Solution{
public:
	int lengthOfLastWord(string s) {
		int n = s.length();
		int i = 0;
		int len = 0;
		if ((n == 1) && (s[0] != ' ')) return 1;
		while (s[i]){
			if (s[i] != ' ') //碰到非空格字符，len++
				len++;
			else if (s[i + 1] && s[i] == ' '&&s[i + 1] != ' ') //碰到空格字符且后一个字符为非空格字符，len清零
				len = 0;
			i++;
		}
		return len;
	}
};

方法二：反序遍历，从字符串末尾开始遍历，碰到非空格字符就开始len++，当碰到非空格字符，且前一个字符为空格字符时，立即结束，返回len

class Solution{
public:
	int lengthOfLastWord(string s) {
		int n = s.length();
		int j = n-1;
		int len = 0;

		while (j>=0){
			if (s[j] != ' ') //碰到非空格字符，len++
				len++;
			if (s[j] != ' '&&s[j - 1] == ' ') //碰到非空格字符，且前一个字符是空格字符，则立即返回len,结束
				return len;
			j--;
		}
		return len;
	}
};

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

class Solution{
public:
	int lengthOfLastWord(string s) {
		int n = s.length();
		int j = n-1;
		int len = 0;

		while (j>=0){
			if (s[j] != ' ') //碰到非空格字符，len++
				len++;
			if (s[j] != ' '&&s[j - 1] == ' ') //碰到非空格字符，且前一个字符是空格字符，则立即返回len,结束
				return len;
			j--;
		}
		return len;
	}
};

int main()
{
    string s = "hello world";
	Solution sol;
	int t = sol.lengthOfLastWord(s);
	cout << t << endl;
	system("pause");
	return 0;  
}