问题：

In the world of Dota2, there are two parties: the Radiant and the Dire.

The Dota2 senate consists of senators coming from two parties. Now the senate wants to make a decision about a change in the Dota2 game. The voting for this change is a round-based procedure. In each round, each senator can exercise one of the two rights:

1. Ban one senator's right: 
   A senator can make another senator lose all his rights in this and all the following rounds.
2. Announce the victory: 
   If this senator found the senators who still have rights to vote are all from the same party, he can announce the victory and make the decision about the change in the game.

Given a string representing each senator's party belonging. The character 'R' and 'D' represent the Radiant party and the Dire party respectively. Then if there are n senators, the size of the given string will be n.

The round-based procedure starts from the first senator to the last senator in the given order. This procedure will last until the end of voting. All the senators who have lost their rights will be skipped during the procedure.

Suppose every senator is smart enough and will play the best strategy for his own party, you need to predict which party will finally announce the victory and make the change in the Dota2 game. The output should be Radiant or Dire.

Example 1:

Input: "RD"
Output: "Radiant"
Explanation: The first senator comes from Radiant and he can just ban the next senator's right in the round 1.
And the second senator can't exercise any rights any more since his right has been banned.
And in the round 2, the first senator can just announce the victory since he is the only guy in the senate who can vote.

Example 2:

Input: "RDD"
Output: "Dire"
Explanation:
The first senator comes from Radiant and he can just ban the next senator's right in the round 1.
And the second senator can't exercise any rights anymore since his right has been banned.
And the third senator comes from Dire and he can ban the first senator's right in the round 1.
And in the round 2, the third senator can just announce the victory since he is the only guy in the senate who can vote.

Note:

1. The length of the given string will in the range [1, 10,000].

解决：

① 模拟了刀塔类游戏开始之前的BP过程，两个阵营按顺序Ban掉对方的英雄，看最后谁剩下来了，就返回哪个阵营。

先统计所有R和D的个数，然后从头开始遍历，如果遇到了R，就扫描环行链之后的所有位置，其实就是坐标对总长度取余，使其不会越界，如果我们找到了下一个D，就将其标记为B，然后对应的计数器cntR自减1。对于D也是同样处理，我们的while循环的条件是cntR和cntD都要大于0，当有一个等于0了的话，那么退出循环，返回那个不为0的阵营即可。

class Solution { //750ms
    public String predictPartyVictory(String senate) {
        int len = senate.length();
        int countR = 0;
        int countD = 0;
        for (char c : senate.toCharArray()){
            if (c == 'R'){
                countR ++;
            }else {
                countD ++;
            }
        }
        if (countR == 0) return "Dire";
        if (countD == 0) return "Radiant";
        char[] schar = senate.toCharArray();
        while(countR > 0 && countD > 0){
            for (int i = 0;i < len;i ++){
                if (schar[i] == 'R'){
                    for (int j = i + 1;j < i + len;j ++){
                        if (schar[j % len] == 'D'){
                            schar[j % len] = 'B';
                            countD --;
                            break;
                        }
                    }
                }else if (schar[i] == 'D'){
                    for (int j = i + 1;j < i + len;j ++){
                        if (schar[j % len] == 'R'){
                            schar[j % len] = 'B';
                            countR --;
                            break;
                        }
                    }
                }
            }
        }
        return countR == 0 ? "Dire" : "Radiant";
    }
}

② 使用两个队列保存两个阵营各自的位置。

class Solution { //24ms
    public String predictPartyVictory(String senate) {
        int len = senate.length();
        Queue<Integer> q1 = new LinkedList<>();
        Queue<Integer> q2 = new LinkedList<>();
        for (int i = 0;i < len;i ++){
            if (senate.charAt(i) == 'R'){
                q1.offer(i);
            }else{
                q2.offer(i);
            }
        }
        while(! q1.isEmpty() && ! q2.isEmpty()){
            int i = q1.poll();
            int j = q2.poll();
            if (i < j){
                q1.offer(i + len);
            }else {
                q2.offer(j + len);
            }
        }
        return q1.size() > q2.size() ? "Radiant" : "Dire";
    }
}

③ 快慢指针

class Solution { //12ms
    public String predictPartyVictory(String senate) {
        char[] schar = senate.toCharArray();
        int len = schar.length;
        int currR = 0;
        int currD = 0;
        while(len > 0){
            int slow = 0;
            int fast = 0;
            int oldLen = len;
            while(fast < len){
                if(schar[fast] == 'D'){
                    if(currR > 0){
                        currR --;
                    }else{
                        currD ++;
                        schar[slow ++] = 'D';
                    }
                }else{
                    if(currD > 0){
                        currD --;
                    }else{
                        currR ++;
                        schar[slow ++] = 'R';
                    }
                }
                fast ++;
            }
            len = slow;
            if(len == oldLen)return schar[0] == 'D' ? "Dire" : "Radiant";
        }
        return "";
    }
}