LEETCODE-Length of Last Word

原创于 2015-10-23 21:18:06 发布 · 285 阅读

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本文介绍了一种计算给定字符串中最后一个单词长度的方法。利用isspace函数定位空格位置，以此确定最后一个单词的起始与结束位置。文章通过具体实例进行说明。

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.
利用了isspace来判断‘空格’的位置；

class Solution {
public:
    int lengthOfLastWord(string s) {
        int n = s.size();
        if(n == 0)
            return 0;
        int lastspace = 0;
        int lastalph = 0;
        for(int i = 0; i < n; i++){
            if(isspace(s[i]))
               lastspace = i+1;
            else
                lastalph = i+1;
        }
        if(lastspace > lastalph){
            lastspace = 0;
            n = lastalph;
            for(int j = 0; j < (lastalph - 1); j++){
                if(isspace(s[j]))
                    lastspace = j + 1;
            }
        }
        if(lastspace == 0)
          return n;
        if(lastspace  > 0 && lastalph > lastspace)
          return n-lastspace;
    }
};