[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、参考别人的，很巧的解法
2、插入排序，排序后，nums⌊ n/2 ⌋必定是出现超过⌊ n/2 ⌋次的数据，而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，只不过速度稍快 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];}