Leetcode22: Length of Last Word

标签：leetcode   algorithm   

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.

一开始想得比较简单，找到最后一个‘ ‘的位置，用s的长度减去这个位置的索引就是最后一个单词的长度。

class Solution {
public:
    int lengthOfLastWord(string s) {
        if(s.empty())
            return 0;
        int index = s.find_last_of(' ');
        return s.length() - index-1;
    }
};

class Solution {
public:
    int lengthOfLastWord(string s) {
        int length = s.size();
        while(s[length-1] == ' ' )
        {
            length--;
        }
        string s1 = s.substr(0, length);
        int index = s1.find_last_of(' ');
        return length-1-index;
    }
};

class Solution {
public:
    int lengthOfLastWord(string s) {
        int length = s.size();
        int sum = 0;
        while(s[length-1] == ' ' )
        {
            length--;
        }
        for(int i = length-1; i >= 0; i--)
        {
            if(s[i] != ' ')
            {
                sum++;
            }
            else
            break;
        }
        return sum;
    }
};

Leetcode22: Length of Last Word

标签：leetcode   algorithm   

原文地址：http://blog.csdn.net/u013089961/article/details/45540361