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  • python(5)

    data = [10, 4, 33, 21, 54, 3, 8, 11, 5, 22, 2, 1, 17, 13]
    
    '''
    思路:有多少个元素就循环多少次,每次循环从第一个元素开始与它后面的元素比较,如果比后面的大,就交换,第次循环最大的数就会被放在最后,那下一次循环就少遍历一位数。
    '''
    
    for j in range(len(data)-1):
        for i in range(len(data)-1-j):
            if data[i] > data[i+1]:
                 data[i], data[i+1] = data[i+1], data[i]
    print(data)

    把代码修改一下,让它打印出每次循环后的结果

    data = [10, 4, 33, 21, 54, 3, 8, 11, 5, 22, 2, 1, 17, 13]
    
    '''
    思路:有多少个元素就循环多少次,从第一个元素开始与它后面的元素比较,如果比后面的大,就交换
    '''
    
    for j in range(len(data)-1):
        for i in range(len(data)-1-j):
            if data[i] > data[i+1]:
                 data[i], data[i+1] = data[i+1], data[i]
        print(data)

    打印

    C:	emp>python3 test.py
    [4, 10, 21, 33, 3, 8, 11, 5, 22, 2, 1, 17, 13, 54]
    [4, 10, 21, 3, 8, 11, 5, 22, 2, 1, 17, 13, 33, 54]
    [4, 10, 3, 8, 11, 5, 21, 2, 1, 17, 13, 22, 33, 54]
    [4, 3, 8, 10, 5, 11, 2, 1, 17, 13, 21, 22, 33, 54]
    [3, 4, 8, 5, 10, 2, 1, 11, 13, 17, 21, 22, 33, 54]
    [3, 4, 5, 8, 2, 1, 10, 11, 13, 17, 21, 22, 33, 54]
    [3, 4, 5, 2, 1, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [3, 4, 2, 1, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [3, 2, 1, 4, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [2, 1, 3, 4, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [1, 2, 3, 4, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [1, 2, 3, 4, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
    [1, 2, 3, 4, 5, 8, 10, 11, 13, 17, 21, 22, 33, 54]
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  • 原文地址:https://www.cnblogs.com/huangxm/p/5274136.html
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