[leetcode]length-of-last-word

RT

思路：

遇到空格就置计数器为0，遇到非空格就+1，尤其注意“a ”，这种情况，所以设置一个 trash变量，存取上一次被丢弃的计数，如果最后以“ ”结束，就从trash中取出计数。

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solution 1

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        int lastWordLength = 0;
        int lastWordLength_trash = lastWordLength; //initialize in case of cases like "   "
        char last = ' ';
        int i=0;
        while(s[i]!='\0'){
            char temp = s[i];
            if(temp!=' '){
                lastWordLength++;    
            //}else if(last!=' '&&temp==' '){
            else if(last!=' '){ //比上面这一行更简洁。
                lastWordLength_trash = lastWordLength;
                lastWordLength=0;
            }
            i++;
            last = temp;
        }
        //avoid cases like "a   "
        if(last==' '){ //get it back from trash
            lastWordLength = lastWordLength_trash;
        }
        return lastWordLength;
    }
};