[Leetcode]-Majority Element

题目：Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times.
找出数组中出现超过⌊ n/2 ⌋次的数据
三种解法：
1、參考别人的。非常巧的解法
2、插入排序，排序后。nums⌊ n/2 ⌋必然是出现超过⌊ n/2 ⌋次的数据。并且插入排序在数据接近排序好的顺序的时候，时间复杂度为O（n）
3、快排，和2一样。仅仅只是速度稍快 12ms

You may assume that the array is non-empty and the majority element always exist in the array.

typedef int elementtype;
//infact CUT = 10 is the best solution
#define CUT 10  
void swap(int *a,int *b)  
{  
    int tem = *a;  
    *a  = *b;  
    *b  = tem;  
}  
void insertion(elementtype A[],int n)  
{  
    int p = 0 ;  
    int j = 0 ;  
    for(p=1;p<n;p++ )  
    {  
        elementtype tem = A[p] ;   
        for(j=p;j>0&&A[j-1]>tem;j--)  
        {  
             A[j] = A[j-1];  
        }  
        A[j] = tem;  
    }  

}  
elementtype median3(elementtype A[],int left ,int right)  
{  
    int center = (left +right) / 2;  
    if(A[left]>A[center])  
        swap(&A[left],&A[center]);  
    if(A[left]>A[right])  
        swap(&A[left],&A[right]);  
    if(A[center]>A[right])  
        swap(&A[center],&A[right]);  

    swap(&A[center],&A[right-1]);  

    return A[right-1];  
}  
void Qsort(elementtype A[],int left, int right)  
{  
    int i,j;  
    elementtype pivot;  

    if(left + CUT<= right)  
    {  
        pivot = median3(A,left,right); //select middle element as pivot  
        i = left;j = right-1;  
        for(;;)  
        {  
            while(A[++i]<pivot){}  

            while(A[--j]>pivot){}  
            if(i<j)  
                swap(&A[i],&A[j]);  
            else  
                break;  
        }  
        swap(&A[i],&A[right-1]);  

        Qsort(A,left,i-1);  
        Qsort(A,i+1,right);  
    }  
    else  
        insertion(A+left,right-left+1);  
}  

int majorityElement(int* nums, int numsSize) {

    //solution 1 :13ms
    /*
    int cnt = 0, res;
    for (int i = 0; i < numsSize; ++i) {
        if (cnt == 0) res = nums[i];
        if (res == nums[i]) ++cnt;
        else --cnt;
    }
    return res;
    */

    //solution 2 insertion sort :28ms
    /*
    int p = 0 ;  
    int j = 0 ;  
    for(p=1;p<numsSize;p++ )  
    {  
        int tem = nums[p] ;   
        for(j=p;j>0&&nums[j-1]>tem;j--)  
        {  
             nums[j] = nums[j-1];  
        }  
        nums[j] = tem;  
    }  
    return nums[numsSize/2];
    */

    //solution 3 qiuk sort :12ms
    Qsort(nums,0,numsSize-1);  
    return nums[numsSize/2];

}