求前K个最大的数

本文介绍3种方法求前K个最大的数。

方法1 排序后求解

排序后求解，这种方法的复杂度是O(nlogn + k)

方法2 优先队列

使用优先队列（说实话，没太明白啥意思），但是我猜大致意思就是说用一个队列来保留当前前K个最大的值。

用堆排序的话，可以只找到前K个最大的数即可。

def findKthLargest(nums, k):
    """
    :type nums: List[int]
    :type k: int
    :rtype: int
    """
    def adjust_heap(nums, end, begin):
        root = begin
        while True:
            tmp = 2 * root + 1
            if tmp > end:
                break
            if tmp+1 <= end  and nums[tmp+1] > nums[tmp]:
                tmp = tmp + 1
            if nums[root] < nums[tmp]:
                nums[root], nums[tmp] = nums[tmp], nums[root]
                root = tmp
            else:
                break

    end = len(nums) - 1
    if end == 0:
        return nums[0]
    begin = end//2 - 1
    for i in xrange(begin, -1, -1):
        adjust_heap(nums, end, i)
    print nums
    for j in xrange(end, end-k, -1):
        nums[j], nums[0] = nums[0], nums[j]
        adjust_heap(nums, j, 0)
        print nums
    return nums[-k]

if __name__ == "__main__":
    nums = [3, 2, 1, 5, 6, 4]
    k = 2
    print findKthLargest(nums, k)

方法3 利用快排的思想

def partition(nums, left, right):
    pviot = nums[right]
    p = left
    for i in xrange(left, right):
        # 由于本题是找第K大的数，因此这里是>
        if nums[i] > pviot:
            nums[i], nums[p] = nums[p], nums[i]
            p += 1
    nums[p], nums[right] = nums[right], nums[p]
    return p

def findKthLargest2(nums, k):
    if len(nums) == 1:
        return nums[0]
    left = 0
    right = len(nums) - 1
    p = partition(nums, left, right)
    while left < right:
        if k == p + 1:
            return nums[p]
        elif k < p + 1:
            return findKthLargest2(nums[left:p], k)
        else:
            return findKthLargest2(nums[p+1:right+1], k-p-1)

def findKthLargest3(nums, k):
    if len(nums) == 1:
        return nums[0]
    size = len(nums)
    left = 0
    right = len(nums) - 1
    while left <= right:
        p = partition(nums, left, right)
        if k == p + 1
            return nums[p]
        elif k > p + 1:
            left = p + 1
        else:
            right = p - 1


if __name__ == "__main__":
    nums = [3, 2, 1, 5, 6, 4]
    k = 2
    print findKthLargest2(nums, k)