python为什么DFS中append到ans的结果不是预期的独立的列表？

Question

求教做了一道OJ题目,调试发现ans的格式是ans: [[[0, 0], [0, 1], [0, 2], [0, 3]], [[0, 0], [0, 1], [0, 2], [1, 2]]]
为什么明明每次执行dfs的都是同级别的对path的append操作,会产生两个大的list.
和预期的每个都是独立元素的list不一致ans: [[0, 0], [0, 1], [0, 2], [0, 3], [0, 0], [0, 1], [0, 2], [1, 2]]

IR = [(0, 1), (1, 0), (0, -1), (-1, 0)]
# input
n = int(input())
code = list(map(int, input().split()))
m = int(input())
book = []
for i in range(m):
    book.append(list(map(int, input().split())))
ans = []
path = []
visited = [[False] * m for i in range(m)]

#dfs
def dfs(code, index, path, visited, ans):
    if index == len(code):
        # ans.append("".join(map(str, path))
        ans.append(path[:])
        return
    for direction in DIR:
        x, y = path[-1][0] + direction[0], path[-1][1] + direction[1]
        if 0 <= x < m and 0 <= y < m and not visited[x][y] and book[x][y] == code[index]:
            path.append([x, y])
            visited[x][y] = True
            dfs(code, index + 1, path, visited, ans)
            path.pop()
            visited[x][y] = False

#map track
for i in range(m):
    for j in range(m):
        if book[i][j] == code[0]:
            path.append([i, j])
            visited[i][j] = True
            dfs(code, 1, path, visited, ans)
            path.pop()
            visited[i][j] = False
#ans format
if len(ans) == 0:
    print("error")
else:
    res = []
    for i in range(len(ans)):
        new_list = []
        for j in range(len(ans[i])):
            new_list+=ans[i][j]
            res.append(new_list)
    res.sort()
    print(" ".join(str(i) for i in res[0]))

因为预期不同 所以改变了format部分 单纯是没想到为什么是这个格式

Answer 1

ans.append(path[:]) -> ans.extend(path)

append 会把 path[:] 当成一个元素插入进去。extend 会把 path 中的每一个元素分别插入进去。差了一层。