911. Online Election

In an election, the i-th vote was cast for persons[i] at time times[i].

Now, we would like to implement the following query function: TopVotedCandidate.q(int t) will return the number of the person that was leading the election at time t.

Votes cast at time t will count towards our query.  In the case of a tie, the most recent vote (among tied candidates) wins.

Example 1:

Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
Output: [null,0,1,1,0,0,1]
Explanation:
At time 3, the votes are [0], and 0 is leading.
At time 12, the votes are [0,1,1], and 1 is leading.
At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
This continues for 3 more queries at time 15, 24, and 8.

Note:

  1. 1 <= persons.length = times.length <= 5000

  2. 0 <= persons[i] <= persons.length

  3. times is a strictly increasing array with all elements in [0, 10^9].

  4. TopVotedCandidate.q is called at most 10000 times per test case.

  5. TopVotedCandidate.q(int t) is always called with t >= times[0].

    这道题是关于线上选举的问题,这年头感觉选举都是得网络者得天下啊,很多都是先在网上形成了一股潮流,比如美国的特朗普,英国的约翰逊,台湾的韩国瑜等等,感觉各个都是社交媒体上的红人,不走寻常路啊。扯远了,拉回本题,其实刚开始博主看了几遍题目,愣是没理解题意,于是去论坛上逛逛,发现也有好多人不清楚,心里稍微舒坦点。这里给了两个数组 persons 和 times,表示在某个时间点 times[i],i这个人把选票投给了 persons[i],现在有一个q函数,输入时间点t,让返回在时间点t时得票最多的人,当得票数相等时,返回最近得票的人。因为查询需求的时间点是任意的,在某个查询时间点可能并没有投票发生,但需要知道当前的票王,当然最傻的办法就是每次都从开头统计到当前时间点,找出票王,但这种方法大概率会超时,正确的方法实际上是要在某个投票的时间点,都统计出当前的票王,然后在查询的时候,查找刚好大于查询时间点的下一个投票时间点,返回前一个时间点的票王即可,所以这里可以使用一个 TreeMap 来建立投票时间点和当前票王之间的映射。如何统计每个投票时间点的票王呢,可以使用一个 count 数组,其中 count[i] 就表示当前i获得的票数,还需要一个变量 lead,表示当前的票王。现在就可以开始遍历所有的投票了,对于每个投票,将票数加到 count 中对应的人身上,然后跟 lead 比较,若当前人的票数大于等于 lead 的票数,则 lead 更换为当前人,同时建立当前时间点和票王之间的映射。在查询的时候,由于时间点是有序的,所以可以使用二分搜索法,由于使用的是 TreeMap,具有自动排序的功能,可以直接用 upper_bound 来查找第一个比t大的投票时间,然后再返回上一个投票时间点的票王即可,参见代码如下:
    解法一:

    class TopVotedCandidate {
    public:
    TopVotedCandidate(vector& persons, vector& times) {

       int n = persons.size(), lead = 0;
       vector<int> count(n + 1);
       for (int i = 0; i < n; ++i) {
           if (++count[persons[i]] >= count[lead]) {
               lead = persons[i];
           }
           m[times[i]] = lead;
       }
    

    }
    int q(int t) {

       return (--m.upper_bound(t))->second;
    

    }

    private:
    map<int, int> m;
    };

    我们也可以用 HashMap 来取代 TreeMap,但因为 HashMap 无法进行时间点的排序,不好使用二分搜索法了,所以就需要记录投票时间数组 times,保存在一个私有变量中。在查询函数中自己来写二分搜索法,是博主之前的总结帖 LeetCode Binary Search Summary 二分搜索法小结 中的第三类,查找第一个大于目标值的数。由于要返回上一个投票时间点,所以要记得减1,参见代码如下:
    解法二:

    class TopVotedCandidate {
    public:
    TopVotedCandidate(vector& persons, vector& times) {

       int n = persons.size(), lead = 0;
       vector<int> count(n + 1);
       this->times = times;
       for (int i = 0; i < n; ++i) {
           if (++count[persons[i]] >= count[lead]) {
               lead = persons[i];
           }
           m[times[i]] = lead;
       }
    

    }
    int q(int t) {

       int left = 0, right = times.size();
       while (left < right) {
           int mid = left + (right - left) / 2;
           if (times[mid] <= t) left = mid + 1;
           else right = mid;
       }
       return m[times[right - 1]];
    

    }

    private:
    unordered_map<int, int> m;
    vector times;
    };

    Github 同步地址:

https://github.com/grandyang/leetcode/issues/911

参考资料:

https://leetcode.com/problems/online-election/

https://leetcode.com/problems/online-election/discuss/173382/C%2B%2BJavaPython-Binary-Search-in-Times

https://leetcode.com/problems/online-election/discuss/191898/Anybody-has-a-magic-general-formula-for-Binary-Search

LeetCode All in One 题目讲解汇总(持续更新中…)


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