Task 3：排序 + 二分查找 (3天)

一、 【排序】
实现归并排序、快速排序、插入排序、冒泡排序、选择排序、堆排序（完成leetcode上的返回滑动窗口中的最大值(239)，这是上一期第三天的任务进行保留（涉及队列可以对第二天进行整理复习））

归并排序：
def merge(a, b):
    c = []
    h = j = 0
    while j < len(a) and h < len(b):
        if a[j] < b[h]:
            c.append(a[j])
            j += 1
        else:
            c.append(b[h])
            h += 1

    if j == len(a):
        for i in b[h:]:
            c.append(i)
    else:
        for i in a[j:]:
            c.append(i)

    return c


def merge_sort(lists):
    if len(lists) <= 1:
        return lists
    middle = len(lists)/2
    left = merge_sort(lists[:middle])
    right = merge_sort(lists[middle:])
    return merge(left, right)


if __name__ == '__main__':
    a = [4, 7, 8, 3, 5, 9]
    print merge_sort(a)

快速排序：
data = [45,3,2,6,3,78,5,44,22,65,46]
 
def quickSort(data, start, end):
    i = start
    j = end
    # i与j重合时，一次排序结束
    if i >= j:
        return
    # 设置最左边的数为基准值
    flag = data[start]
    while i < j:
        while i<j and data[j] >= flag:
            j -= 1
        # 找到右边第一个小于基准的数，赋值给左边i。此时左边i被记录在flag中
        data[i] = data[j]
        while i<j and data[i] <= flag:
            i += 1
        # 找到左边第一个大于基准的数，赋值给右边的j。右边的j的值和上面左边的i的值相同
        data[j] = data[i]
    # 由于循环以i结尾，循环完毕后把flag值放到i所在位置。
    data[i] = flag
    # 除去i之外两段递归
    quickSort(data, start, i-1)
    quickSort(data, i+1, end)
 
quickSort(data,0, len(data)-1)
print data

插入排序：
def insert_sort(ary):
    n = len(ary)
    for i in range(1,n):
        if ary[i] < ary[i-1]:
            temp = ary[i]
            index = i           #待插入的下标
            for j in range(i-1,-1,-1):  #从i-1 循环到 0 (包括0)
                if ary[j] > temp :
                    ary[j+1] = ary[j]
                    index = j   #记录待插入下标
                else :
                    break
            ary[index] = temp
    return ary

冒泡排序：
def bubble_sort(list_1):
    n = len(list_1)
    for j in range(n-1):
        for i in range(0,n-1-j):
            if list_1[i]>list_1[i+1]:
                list_1[i],list_1[i+1] = list_1[i+1],list_1[i]

if __name__ == '__main__':
    list_1 = [2,23,42,1,9]
    bubble_sort(list_1)
    print(list_1)
 输出结果：
 [1, 2, 9, 23, 42]

选择排序：
#selectSort which is a time-consuming sort algorithm.Its Time-complexity is O(N**2)
#1、we just use the simple sort algorithm to get the smallest number per loop,which the time-consuming is O(n)
#2、next we can define a function that a loop to get the small value every loop,and the append the value into a new Array
#to get the minmum value
def getmin(arr):
    min = arr[0];
    min_index = 0;
    for i in range(0,len(arr)):
        if arr[i]<min:
            min = arr[i]
            min_index = i
    return min_index

#SelectSort
def selectSort(arr):
    newArr = [];
    for i in range(0,len(arr)):
        min = getmin(arr);
        newArr.append(arr.pop(min))
    return newArr;

#test the output
a = [4,6,9,1,3,87,41,5]
print(selectSort(a))

堆排序：
from collections import deque


def swap_param(L, i, j):
    L[i], L[j] = L[j], L[i]
    return L


def heap_adjust(L, start, end):
    temp = L[start]

    i = start
    j = 2 * i

    while j <= end:
        if (j < end) and (L[j] < L[j + 1]):
            j += 1
        if temp < L[j]:
            L[i] = L[j]
            i = j
            j = 2 * i
        else:
            break
    L[i] = temp


def heap_sort(L):
    L_length = len(L) - 1

    first_sort_count = L_length / 2
    for i in range(first_sort_count):
        heap_adjust(L, first_sort_count - i, L_length)

    for i in range(L_length - 1):
        L = swap_param(L, 1, L_length - i)
        heap_adjust(L, 1, L_length - i - 1)

    return [L[i] for i in range(1, len(L))]


def main():
    L = deque([50, 16, 30, 10, 60,  90,  2, 80, 70])
    L.appendleft(0)
    print heap_sort(L)


if __name__ == '__main__':
    main()
这个代码没有跑出结果，报错！待修改

二、 编程实现 O(n) 时间复杂度内找到一组数据的第 K 大元素
【二分查找】
实现一个有序数组的二分查找算法
实现模糊二分查找算法（比如大于等于给定值的第一个元素）

来源：https://blog.csdn.net/weixin_43148174/article/details/99337944