数据结构作业详解：最大子列和问题及解决方法-优快云博客

本文详细介绍了数据结构课程中关于最大子列和问题的两个典型作业题，包括问题背景、输入输出格式、解题思路与代码实现。通过实例解析，帮助读者理解并掌握如何解决此类问题，特别强调了动态规划的思想应用。

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中国大学MOOC。数据结构。作业。

这门课作业不算成绩。老师说可以随便交流。

小白写个作业也不容易，就存一份。

比较好的算法如下图。

01-1. 最大子列和问题(20)

给定K个整数组成的序列{ N1, N2, ..., NK }，“连续子列”被定义为{ Ni, Ni+1, ..., Nj }，其中 1 <= i <= j <= K。“最大子列和”则被定义为所有连续子列元素的和中最大者。例如给定序列{ -2, 11, -4, 13, -5, -2 }，其连续子列{ 11, -4, 13 }有最大的和20。现要求你编写程序，计算给定整数序列的最大子列和。

输入格式：

输入第1行给出正整数 K (<= 100000)；第2行给出K个整数，其间以空格分隔。

输出格式：

在一行中输出最大子列和。如果序列中所有整数皆为负数，则输出0。

输入样例：

6
-2 11 -4 13 -5 -2

输出样例：

20

第一题就按老师视频里（上面那张图）抄下来就行了。秒过。

主要思想是：ThisSum一直从左往右累加，如果比原来的最大值MaxSum大就更新最大值，如果为负数就清零；最后返回最大值。

amount = raw_input()
num_list = [int(i) for i in raw_input().split(' ')]

ThisSum = 0; MaxSum = 0

for i in num_list:
    ThisSum += i
    if ThisSum > MaxSum:
        MaxSum = ThisSum
    elif ThisSum < 0:
        ThisSum = 0

print MaxSum

01-2. Maximum Subsequence Sum (25)

Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.
Sample Input:

10
-10 1 2 3 4 -5 -23 3 7 -21

Sample Output:

10 1 4

第二题搞了好几个小时。

主要思想：最大值同上一题，在for循环中用#号标出，不要动它；最后一个数字显然就是ThisSum大于MaxSum时候的那个数；那么第一个数字怎么来？这里用的step来减。每过一个数，step加一，如果ThisSum此时为负就和它一起清零。

另外，测试点5特别不好过，刷了讨论区才知道这个是由负数和零组成的，自己举了下面第三个例子，加了三个if，看上去很丑不过终于过了！

amount = int(raw_input())
num_list = [int(i) for i in raw_input().split(' ')]

# amount = 10
# num_list = [-4,2,5,-9,1,3,4,0,-4,0] # 8 1 4

# amount = 10
# num_list = [-10,1,2,3,4,-5,-23,3,7,-21] # 10 1 4

# amount = 5
# num_list =[-8,0,-2,0,-1] #0 0 0

ThisSum = 0; MaxSum = 0
first = 0; last = -1; step = 0
count = 0; flag = False

for i in range(amount):
    ThisSum += num_list[i] #
    step += 1
    if (num_list[i] <= 0):
        count += 1
        if (num_list[i] == 0): flag = True
    if (ThisSum > MaxSum):
        MaxSum = ThisSum #
        first = i - step + 1
        last = i
    elif (ThisSum < 0):
        ThisSum = 0 #
        step = 0

FirstNum = num_list[first]; LastNum = num_list[last]

if count == len(num_list) and flag: FirstNum = 0; LastNum = 0

print MaxSum, FirstNum, LastNum