数据结构练习——最大子列和问题python实现

网易云课堂《数据结构》最大子列和问题的python实现(地址：http://mooc.study.163.com/learn/1000033001?tid=2001462004#/learn/content?type=detail&id=2001778303)

首先声明create_sequence(num)函数来创建一个长度为num的list。

1.def create_sequence(num):  
2.    i = 0  
3.    sequence = []  
4.    while i < num:  
5.        sequence.append(random.randrange(-20,20))  
6.        i += 1  
7.    return sequence,num  

算法1：粗暴的比较所有子列的和来找到最大值,算法复杂度O(n3)

1.#算法1遍历所有子列后记录最大子列  
2.def algorithm1(sequence,num):  
3.    start_time = time.clock()  
4.    thisSum = 0 #记录当前子列和  
5.    maxSum = 0  #记录最大子列和  
6.    i = 0  
7.    j = 0  
8.    k = 0  
9.    while i < num:  
10.        j = i  
11.        while j < num:  
12.            thisSum = 0  
13.            k = i  
14.            while k <= j:  
15.                thisSum += sequence[k]  
16.                k += 1  
17.            if thisSum > maxSum:  
18.                maxSum = thisSum  
19.            j += 1  
20.        i += 1  
21.    end_time = time.clock()  
22.    return maxSum,end_time-start_time  

算法2：针对算法1进行的改进，将三重循环减至二重循环，算法复杂度0(n2)

1.def algorithm2(sequence,num):  
2.    start_time = time.clock()  
3.    thisSum = 0  
4.    maxSum = 0  
5.    i = 0  
6.    j = 0  
7.    while i < num :  
8.        thisSum = 0  
9.        j = i  
10.        while j < num: #对于相同的i,不同的j,只要在j-1次循环的基础上累加一项即可，这样可以减少一轮迭代，降低时间复杂度  
11.            thisSum += sequence[j]  
12.            if thisSum > maxSum:  
13.                maxSum = thisSum  
14.            j += 1  
15.        i += 1  
16.    end_time = time.clock()  
17.    return maxSum,end_time-start_time 

算法3：在线处理，算法复杂度O(N)

1.def algorithm4(sequence,num):  
2.    start_time = time.clock()  
3.    thisSum = 0  
4.    maxSum = 0  
5.    i = 0  
6.    while i < num:  
7.        thisSum += sequence[i]  #向右累加  
8.        if thisSum > maxSum:  
9.            maxSum = thisSum    #发现更大和则更新当前的结果  
10.        elif thisSum < 0:       #如果当前子列和为负，则不可能使后面的部分和增大，就可以抛弃  
11.            thisSum = 0  
12.        i += 1  
13.    if maxSum == 0:             #考虑list中各项都不是正数的情况  
14.        tempSequence = sequence.sort(reverse=True)  
15.        if 0 in sequence:       #如果存在0，则最大值为0  
16.            return 0  
17.        else:                   #否则子列的最大值就为最大的负值  
18.            return sequence[0]  
19.    end_time = time.clock()  
20.    return maxSum,end_time-start_time  

测试代码和运行结果：

1.if __name__ == '__main__':  
2.    sequence,num = create_sequence(10)  
3.    result,time_1_1 = algorithm1(sequence,num)  
4.    result,time_2_1 = algorithm2(sequence,num)  
5.    result,time_3_1 = algorithm4(sequence,num)  
6.    print('当list长度为10时，运行时间分别为：',time_1_1,time_2_1,time_3_1)  
7.  
8.    print('########################')  
9.    sequence,num = create_sequence(100)  
10.    result,time_1_2 = algorithm1(sequence,num)  
11.    result,time_2_2 = algorithm2(sequence,num)  
12.    result,time_3_2 = algorithm4(sequence,num)  
13.    print('当list长度为100时，运行时间分别为：',time_1_2,time_2_2,time_3_2)  
14.  
15.    print('########################')  
16.    sequence,num = create_sequence(1000)  
17.    result,time_1_3 = algorithm1(sequence,num)  
18.    result,time_2_3 = algorithm2(sequence,num)  
19.    result,time_3_3 = algorithm4(sequence,num)  
20.    print('当list长度为100时，运行时间分别为：',time_1_3,time_2_3,time_3_3)  

1.当list长度为10时，运行时间分别为： 7.943201645759982e-05 2.1419869606543776e-05 5.3549674016359475e-06  
2.########################  
3.当list长度为100时，运行时间分别为： 0.030387208767866623 0.0015895328237189449 2.5436095157760352e-05  
4.########################  
5.当list长度为100时，运行时间分别为： 33.7867263745532 0.12402506124743695 0.0002463285004736804  

视频中还介绍了一种分而治之的算法暂未实现。