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",
return5.

解法一：

順序遍歷，有單詞則len++，有空格且下一位為單詞則len=0，直到末尾輸出len

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        int len=0;
        while(*s)
        {
            if(*s++!=‘ ‘)
                len++;
            else if(*s&&*s!=‘ ‘)
                len=0;
        }
        return len;
    }
};
 

解法二：

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        int count = 0;
        int len = strlen(s);
//反向查找，末尾空格忽略，行中出現空格就終止循環
        for(int i = len-1; i >= 0 ; i--){
            if(s[i] == ‘ ‘){
                if(count)
                    break;
            }
            
 
else{
                count++;
            }
        }
        return count;
    }
};

length-of-last-word 最後一個單詞的長度