2020年 Moore majority vote algori

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第一眼看到这个题目，想到的是使用Map来统计出现频次，然后遍历找出频次大于n/2的元素。

class Solution {
    public int majorityElement(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        for(Integer item: nums){
            if(null != map.get(item)){
                map.put(item, map.get(item) + 1);
            }else
                map.put(item,1);
        }
        for(Integer freq: map.keySet()){
            if(map.get(freq) > (nums.length / 2))
                return freq;
        }
        return -1;
    }
}

但是显而易见的是：没有用上n/2这个设定。
其实这个看似简单的问题有一个特殊的解法，就是我们要说的Boyer–Moore majority vote algorithm ：摩尔投票法
知乎上有老哥给出了非常形象的解释：

核心是对拼消耗：类似我们玩的即时战略游戏：魔兽争霸，三国群英传等。假设地图上有一家（称作红色军）拥有所有军队中一半以上的小兵，在直接对拼下不虚任何对手（不同队伍小兵1v1地同归于尽），其他队伍像蓝色、绿色、紫色等，有可能会互相消耗，但是最后留在地图上的一定是同一队人数最多的红色。

如何理解摩尔投票算法？ - 知乎用户的回答 - 知乎
https://www.zhihu.com/question/49973163/answer/617122734
具体实现：

class Solution{
    public int majorityElement(int[] nums){
    int res = 0;
        int count = 0;

        for (int num : nums) {
            if (count == 0) {
                res = num;
            }
            count += (num == res) ? 1 : -1;
        }

        return res;
    }

有兴趣的话还可以再看一下下面这个
wiki上的伪代码：

The algorithm maintains in its local variables a sequence element and a counter, with the counter initially zero. 
It then processes the elements of the sequence, one at a time. When processing an element x, if the counter is zero, the algorithm stores x as its remembered sequence element and sets the counter to one. 
Otherwise, it compares x to the stored element and either increments the counter (if they are equal) or decrements the counter (otherwise). 
At the end of this process, if the sequence has a majority, it will be the element stored by the algorithm. This can be expressed in pseudocode as the following steps:

Initialize an element m and a counter i with i = 0
For each element x of the input sequence:
If i = 0, then assign m = x and i = 1
else if m = x, then assign i = i + 1
else assign i = i − 1
Return m
Even when the input sequence has no majority, the algorithm will report one of the sequence elements as its result. 
However, it is possible to perform a second pass over the same input sequence in order to count the number of times the reported element occurs and determine whether it is actually a majority. This second pass is needed, as it is not possible for a sublinear-space algorithm to determine whether there exists a majority element in a single pass through the input.
            

--- 引用自wikipedia