LeetCode | Length of Last Word(最后一个单词的长度)


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.

题目解析:

让求最后的单词的长度,本来以为只要碰到空格就更新index,最后用字符串长度len与index之间的差值表示就行。但是忘了一种情况“a ”,在单词最后还有空格,那么就需要特殊处理。本来还想用前后指针i,j不过也没得到很好的思路。后来通过计数count来判断。当碰到空格,并且空格的下一个是字符的时候,才将count清零表示重新计算。并且遇到空格还不能让count++。注意这些情况,就能得到正确的结果了。

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        int len = strlen(s);
        if(s == NULL || len == 0)
            return 0;
        int count = 0;
        if(s[0] != ' ')
            count++;
        for(int i = 1;i < len;i++)
            if(s[i] == ' '){
                if(i+1 < len && s[i+1] != ' ')
                    count = 0;
            }else
                count++;

        return count;
    }
};

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