技術標籤：leetcode刷題leetcode

兩個佇列模擬，彷彿麻煩了點。

class Solution {
public:
    string predictPartyVictory(string senate) {
        deque<char> queue;
        deque<char> tmp;
        int D=0,R=0,D_flag=0,R_flag=0;
        for(char &c:senate) queue.push_back(c);

        while(true){
            for
 
(char &c:queue) {
                if(c=='D') {
                    D_flag=1;
                    if(D>=0) tmp.push_back('D');
                    R-=1;
                    D+=1;
                }else{
                    R_flag=1;
                    if(R>=0)  tmp.push_back('R');
                    D-
 
=1;
                    R+=1;
                }
            }
            if(R_flag==0||D_flag==0) break;
            queue = tmp;
            tmp.clear();
            D_flag=R_flag=0;
        }
        return R_flag==0?"Dire":"Radiant";
    }
};

這樣的效率高一些

class Solution {
public:
    string predictPartyVictory
 
(string senate) {
        int n = senate.size();
        queue<int> R,D;
        for(int i=0;i<n;i++) {
            if(senate[i]=='R') R.push(i);
            else D.push(i);
        }
        while(!R.empty()&&!D.empty()) {
            if(R.front()<D.front()) {
                R.push(R.front()+n);
            }else D.push(D.front()+n);
            R.pop();
            D.pop();
        }
        if(R.empty()) return "Dire";
        return "Radiant"; 
    }
};