[Leetcode] 649. Dota2 Senate 解题报告

题目：

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:

Ban one senator's right:
A senator can make another senator lose all his rights in this and all the following rounds.
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:

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

思路：

这是一道贪心算法的题目：对于每个议员而言，最好的策略就是禁止对方党派下一个议员行使权力，所以每次就用贪心算法找到下一个对方党派的议员，并删除。这样不断迭代，直到某一方党派的人全部被删除。在实现中，我们用两个队列来模拟两个党派的人：取出两个队首元素，看看谁行使权力的时间更靠前，然后将行使权力时间靠前的议员从队首移到队尾，将时间靠后的议员直接从队列中删除。如何判断队员行使权力的时间呢？我们可以在队列中用int来表示行使权力的顺序号，每次判断两个队列中int的大小即可。

代码：

class Solution {
public:string predictPartyVictory(string senate) {queue<int> q1, q2;int n = senate.length();for (int i = 0; i < n; ++i) {senate[i] == 'R' ? q1.push(i) : q2.push(i);}while (!q1.empty() && !q2.empty()) {int r_index = q1.front(), d_index = q2.front();q1.pop(), q2.pop();(r_index < d_index) ? q1.push(r_index + n) : q2.push(d_index + n);}return q1.size() > q2.size() ? "Radiant" : "Dire";}
};

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