用分治法求众数（seek the public with the divide and conquer）

// 用分治法求众数


//参考：http://blog.csdn.net/core__code/article/details/47385045
#include <iostream>
#include <cstdio>

using namespace std;

void split(int s[], int n, int &l, int &r)  //here should have a low limit, because the split for the top half will don't need to travel the whole array;
{
    int middle = n / 2;
    for(l=0; l<n; l++)
    {
        if(s[l]==s[middle])
        {
            break;
        }
    }

    for(r=middle+1; r<n; r++)
    {
        if(s[r]!=s[middle])
        {
            break;
        }
    }
    return;
}

void getMaxCount(int s[], int n, int &maxCount, int &targetValue)  //high is not the upper of array, but the contain of array;
{
    int l, r;   //l is the left edge of middle elements, r is right
    split(s, n, l, r);
    int middle = n / 2;
    if(r-l>maxCount)
    {
        maxCount = r - l;
        targetValue = s[middle];
    }
    if(maxCount<l)  //if the number on the left is less than the number of already getting the number of maxcount, the recurrence will not need to continue
    {
        getMaxCount(s, l, maxCount, targetValue);  //l is impossible to arrive
    }
    if(maxCount<n-r)
    {
        getMaxCount(s+r, n-r, maxCount, targetValue);
    }
    return;
}

int main(void)
{
    int s[] = {1, 2, 2, 2, 3, 3, 3, 3, 3, 5, 6, 6, 6, 6, 6, 6};
    int n = sizeof(s)/sizeof(s[0]);
    int maxCount = 0;
    int targetValue = 0;    //this is meaningless, because the maxCount is zero, the showing times of every num is bigger than one
    getMaxCount(s, n, maxCount, targetValue);
    cout << targetValue << " " << maxCount;
    return 0;
}