topk问题的解决思路

topk问题：在n个数中取出前k大的数

实现思路

# 思路1：快排，排序后切片出最大的k个；时间复杂度: O(nlogn) + O(k)   O(k)可忽略不计

# 思路2：冒泡、选择排序，循环k趟，只获取前k个大数； 时间复杂度： O(kn)

# 思路3：插入排序，维护k长度的有序列表，其他元素挨个和这个列表最小的数比较，小的丢弃，大的插入排序进去；时间复杂度：O(kn)

# 思路4：堆排序，建k长度的小根堆，其他元素挨个和堆顶元素比较，大的替换堆顶元素再向下调整为小根堆。最后挨个出数；时间复杂度：O(nlogk)

代码实现

快排实现

def quick_sort(li, left, right):
    if left < right:
        mid = partition(li, left, right)
        quick_sort(li, left, mid-1)
        quick_sort(li, mid+1, right)

def partition(li, left, right):
    tmp = li[left]
    while left < right:
        while left < right and li[right] >= tmp:
            right -= 1
        li[left] = li[right]
        while left < right and li[left] <= tmp:
            left += 1
        li[right] = li[left]
    li[left] = tmp
    return left

# 测试
li = [i for i in range(10)]
random.shuffle(li)
print('排序前',li)
quick_sort(li, 0, len(li)-1)
print('排序后', li)
topk = li[-5:]
print('topk', topk)

# output:
排序前 [7, 2, 1, 6, 5, 4, 0, 3, 9, 8]
排序后 [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
topk [5, 6, 7, 8, 9]

冒泡排序实现

def topk(li, k):
    for i in range(k):
        for j in range(len(li)-1-i):
            if li[j] > li[j+1]:
                li[j], li[j+1] = li[j+1], li[j]
                
# 测试
li = [i for i in range(10)]
random.shuffle(li)
print('排序前',li)
topk(li, 5)
print('排序后', li)
print('topk', li[-5:])

# output:
排序前 [8, 9, 7, 2, 4, 5, 3, 0, 1, 6]
排序后 [2, 3, 0, 1, 4, 5, 6, 7, 8, 9]
topk [5, 6, 7, 8, 9]

选择排序实现

def topk(li, k):
    for i in range(k):
        max_index = i
        for j in range(i+1, len(li)):
            if li[j] > li[max_index]:
                max_index = j
        li[i], li[max_index] = li[max_index], li[i]

# 测试
li = [i for i in range(10)]
random.shuffle(li)
print('排序前',li)
topk(li, 5)
print('排序后', li)
print('topk', li[:5])

# output:
排序前 [6, 9, 0, 3, 1, 7, 5, 4, 8, 2]
排序后 [9, 8, 7, 6, 5, 0, 1, 4, 3, 2]
topk [9, 8, 7, 6, 5]

插入排序实现

# 待补充......

堆排序实现

def sift(li, low, high):
    i = low
    j = 2 * i + 1
    tmp = li[low]

    while j <= high:
        if j+1 <= high and li[j+1] < li[j]:
            j = j + 1
        if li[j] < tmp:
            li[i] = li[j]
            i = j
            j = 2 * i + 1
        else:
            li[i] = tmp
            break
    else:
        li[i] = tmp


def topk(li, k):
    # heap建小根堆
    heap = li[0:k]
    for i in range(k//2-1, -1, -1):
        sift(heap, i, k-1)

    # 遍历将大于heap[0]的元素换掉并向下调整
    for i in range(k, len(li)):
        if li[i] > heap[0]:
            heap[0] = li[i]
            sift(heap, 0, k-1)
    # 挨个出数
    for i in range(k-1, -1, -1):
        heap[0], heap[i] = heap[i], heap[0]
        sift(heap, 0, i-1)

    return heap

# 测试
li = [i for i in range(10)]
random.shuffle(li)
print('排序前',li)
heap = topk(li, 5)
print('topk', heap)

# output:
排序前 [9, 1, 0, 6, 2, 4, 7, 3, 8, 5]
topk [9, 8, 7, 6, 5]