增量重复

Question

我有一个整数列表，列表是循环。我想找到连续的重复项，而不是将重复的数字 + 1 放在起始位置。还需要这样做，直到列表中没有连续的重复项或列表为空。我尝试了不同形式的 while 循环，但无法达到我的目标。尝试用 itertools.cycle 循环，因为 -1 和 0 也是连续的，但无法删除或插入。还找出何时停止是问题，尝试了 int_list == list(set(int_list)) 但可能有一些不连续的重复，因此创建了一个无限循环。这是一个例子：

int_list = [1,1,1,2,3,4,5,6,7,7,8,8,9,10,8,11,4,12,15,13,14,14,15,16,1]
processed_list = [11,8,11,4,12,15,13,17]

编辑：

原始问题在 this Codewars 套路中。

Answer 1

deque接近于循环列表，因为它允许旋转列表。我在下面的代码中利用了这一点。

该算法将 item[0] 与下一项 item[1] 以及前一项（循环意义上的）进行比较 item[-1] 丢弃所有重复项并递增原始项 item[0]如果发现任何重复项。

如果没有找到重复项，则旋转列表并使用新的 item[0] 重复该过程。

如果没有找到重复的完整列表轮换，则该过程停止。

# Version 3

from collections import deque

def process(inp):
    circ = deque(inp)
    rcnt = 0    # rotations without finding a dup
    while rcnt < len(circ):
        n = circ[0]
        isdup = False
        while len(circ) >= 1 and circ[1] == n:
            isdup = True
            circ.popleft()
        while len(circ) >= 1 and circ[-1] == n:
            isdup = True
            circ.pop()
        if isdup:
            circ[0] = n+1
            rcnt = 0
        else:
            circ.rotate(1)
            rcnt += 1
        # print(list(circ)) # uncomment to watch the progress
    return list(circ)

test = [1,1,1,2,3,4,5,6,7,7,8,8,9,10,8,11,4,12,15,13,14,14,15,16,1]
print(process(test))

带有未注释的调试 print:

[2, 2, 3, 4, 5, 6, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[3, 3, 4, 5, 6, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[4, 4, 5, 6, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[5, 5, 6, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[6, 6, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[7, 7, 7, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[8, 8, 8, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[9, 9, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[10, 10, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[11, 8, 11, 4, 12, 15, 13, 14, 14, 15, 16]
[16, 11, 8, 11, 4, 12, 15, 13, 14, 14, 15]
[15, 16, 11, 8, 11, 4, 12, 15, 13, 14, 14]
[14, 15, 16, 11, 8, 11, 4, 12, 15, 13, 14]
[15, 15, 16, 11, 8, 11, 4, 12, 15, 13]
[16, 16, 11, 8, 11, 4, 12, 15, 13]
[17, 11, 8, 11, 4, 12, 15, 13]
[13, 17, 11, 8, 11, 4, 12, 15]
[15, 13, 17, 11, 8, 11, 4, 12]
[12, 15, 13, 17, 11, 8, 11, 4]
[4, 12, 15, 13, 17, 11, 8, 11]
[11, 4, 12, 15, 13, 17, 11, 8]
[8, 11, 4, 12, 15, 13, 17, 11]
[11, 8, 11, 4, 12, 15, 13, 17]
[17, 11, 8, 11, 4, 12, 15, 13]
[13, 17, 11, 8, 11, 4, 12, 15]

the last line is the result:
[13, 17, 11, 8, 11, 4, 12, 15]

增量重复

Incremental duplicates

python

list

duplicates

python-3.x