Python 算法之冒泡排序

 冒泡排序

冒泡排序算法的原理如下：（从后往前）

1、比较相邻的元素。如果第一个比第二个大，就交换他们两个。
2、对每一对相邻元素作同样的工作，从开始第一对到结尾的最后一对。在这一点，最后的元素应该会是最大的数。
3、针对所有的元素重复以上的步骤，除了最后一个。
4、持续每次对越来越少的元素重复上面的步骤，直到没有任何一对数字需要比较。　　

对下列数据进行排序

 10,4,33,21,54,3,8,11,5,22,2,1,17,13,6 

代码：

#/usr/bin/eve python
# -*- coding:utf-8 -*-

data = [10,4,33,21,54,3,8,11,5,22,2,1,17,13,6]

l = len(data)

for i in range(l):
    for j in range(l-1-i):
        if data[j] > data[j+1]:
            tmp = data[j]
            data[j] = data[j+1]
            data[j+1] = tmp
    print(data)
print(data)

　　

输出结果：

[4, 10, 21, 33, 3, 8, 11, 5, 22, 2, 1, 17, 13, 6, 54]
[4, 10, 21, 3, 8, 11, 5, 22, 2, 1, 17, 13, 6, 33, 54]
[4, 10, 3, 8, 11, 5, 21, 2, 1, 17, 13, 6, 22, 33, 54]
[4, 3, 8, 10, 5, 11, 2, 1, 17, 13, 6, 21, 22, 33, 54]
[3, 4, 8, 5, 10, 2, 1, 11, 13, 6, 17, 21, 22, 33, 54]
[3, 4, 5, 8, 2, 1, 10, 11, 6, 13, 17, 21, 22, 33, 54]
[3, 4, 5, 2, 1, 8, 10, 6, 11, 13, 17, 21, 22, 33, 54]
[3, 4, 2, 1, 5, 8, 6, 10, 11, 13, 17, 21, 22, 33, 54]
[3, 2, 1, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[2, 1, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]
[1, 2, 3, 4, 5, 6, 8, 10, 11, 13, 17, 21, 22, 33, 54]　　

 生成一个4*4的2维数组并将其顺时针旋转90度

代码：

#!/usr/bin/env python
# -*- conding-utf8 -*-

def xz(sz, n):
    for i in range(n):
        for j in range(i, n):
            if i == j:
                continue
            else:
                t = sz[i][j]
                sz[i][j] = sz[j][i]
                sz[j][i] = t

    return sz

data = [[col for col in range(4)] for row in range(4)]
for i in data:
    print(i)
print("----------------------------")

b = xz(data, 4)
for i in b:
    print(i)　　

输出结果：

[0, 1, 2, 3]
[0, 1, 2, 3]
[0, 1, 2, 3]
[0, 1, 2, 3]
----------------------------
[0, 0, 0, 0]
[1, 1, 1, 1]
[2, 2, 2, 2]
[3, 3, 3, 3]