414. Third Maximum Number

/*

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

*/
//66ms 3.41%
//思路：先把数组排一遍序，用快速排序，然后去重复元素，这些都是我之前写好的leetcode题目中写好的函数，直接拿过来用了。然后输出第三个大数
int thirdMax(int* nums, int numsSize) {
    int length;
    quicksort(nums,0,numsSize-1);
    length = removeDuplicates(nums,numsSize);
    if(length<3)
        return nums[length-1];
    return nums[length-3];
}
void quicksort(int* nums,int left,int right)
{
    int i,j,t,temp;
    if(left>right)
       return;

    temp=nums[left]; //temp中存的就是基准数
    i=left;
    j=right;
    while(i!=j)
    {
                   //顺序很重要，要先从右边开始找
                   while(nums[j]>=temp && i<j)
                            j--;
                   //再找右边的
                   while(nums[i]<=temp && i<j)
                            i++;
                   //交换两个数在数组中的位置
                   if(i<j)
                   {
                            t = nums[i];
                            nums[i] = nums[j];
                            nums[j] = t;
                   }
    }
    //最终将基准数归位
    nums[left] = nums[i];
    nums[i] = temp;

    quicksort(nums,left,i-1);//继续处理左边的，这里是一个递归的过程
    quicksort(nums,i+1,right);//继续处理右边的 ，这里是一个递归的过程
}
int removeDuplicates(int* nums, int numsSize) {
    int i,count = 1;
    if(nums[0]==NULL&&numsSize==0)
        return NULL;
    for(i=0;i<numsSize-1;i++)
        if(nums[i]!=nums[i+1])
            nums[count++] = nums[i+1];
    return count;
}