变量简洁正确完整思路

left来到一个字符，right来到最后一个字符，如果s[left]不是空格就word+=s[left]，遇到空格将单词push_back到双端队列deque ans1，string ans从ans1前部取单词+=并加空格，ans1空了就不要加空格   精确定义 left right ans1 ans word

class Solution {
public:
    string reverseWords(string s) {
        int n=s.size();
        int left=0,right=n-1;
        while(left<n&&s[left]==' ')left++;
        while(right>=0&&s[right]==' ')right--;
        deque<string>ans1;
        while(left<=right){
            //cout<<s[left]<<' '<<s[right]<<endl;
            string word;
            while(s[left]!=' '&&left<=right){
                word+=s[left];
                left++;
            }
            if(word.size()>0)ans1.push_front(word);
            left++;
        }
        string ans;
        for(auto str:ans1){
            ans+=str;
            ans+=' ';
        }
        ans.pop_back();
        return ans;
    }
};

class Solution {
public:
    string reverseWords(string s) {
        int n=s.size();
        int left=0,right=n-1;
        while(left<n&&s[left]==' ')left++;
        while(right>=0&&s[right]==' ')right--;
        deque<string>ans1;
        while(left<=right){
            //cout<<s[left]<<' '<<s[right]<<endl;
            string word;
            while(s[left]!=' '&&left<=right){
                word+=s[left];
                left++;
            }
            if(word.size()>0)ans1.push_front(word);
            left++;
        }
        string ans;
        for(auto str:ans1){
            ans+=str;
            ans+=' ';
        }
        ans.pop_back();
        return ans;
    }
};

 

踩过的坑 这种类型题考察的是debug能力，cout什么是最恰当的？s[left]比left好，积累             if(word.size()>0)ans1.push_front(word);遇到空格，word都没有积累             while(s[left]!=' '&&left<=right){溢出了