<LeetCode OJ> 58. Length of Last Word

58. Length of Last Word

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Question

Total Accepted: 78112 Total
Submissions: 275441 Difficulty: Easy

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

.

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//思路首先：简单模拟思想，去掉尾部的空格并统计是字符的个数直到再次遇到空格就停止统计
class Solution {
public:
    int lengthOfLastWord(string s) {
        int strLen=s.size();
        int len=0;
        int i=0;
        while(s[strLen-i-1]==' ')
            i++;
        for(;i<strLen;i++)
        {
            if(s[strLen-i-1]>='A'&&s[strLen-i-1]<='Z' || s[strLen-i-1]>='a'&&s[strLen-i-1]<='z' )
                len++;
            else
                break;
        }
        return len;
    }
};

或者用STL来写：

class Solution {
public:
    int lengthOfLastWord(string s) {
        int i = s.find_last_not_of(' ');
        return (i == string::npos) ? 0 : (i - s.find_last_of(' ', i));
    }
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50498956

原作者博客：http://blog.csdn.net/ebowtang