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1. BFS
- 111 二叉树的最小深度
- 725 打开转盘锁
- 773 滑动拼图
2. 题目
【111 二叉树的最小深度】
def minDepth(root):'''给定⼀个⼆叉树,找出其最⼩深度,最⼩深度是从根节点到最近叶⼦节点(没有⼦节点的节点)的最短路径上的节点数量leetcode: 111 二叉树的最小深度input: [3,9,20,null,nul, 15, 7]3/ \9 20/ \15 7output: 2思路:1.2.3.'''if root is None:return 0q, depth = [root], 1while len(q) != 0:size = len(q)for i in range(size):cur = q[0]q.remove(q[0])if cur.left is None and cur.right is None:return depthif cur.left is not None:q.append(cur.left)if cur.right is not None:q.append(cur.right)depth += 1return depth
【725 打开转盘锁】
def openLock(deadends, target):'''你有一个带有四个圆形拨轮的转盘锁。每个拨轮都有10个数字: '0', '1', '2', '3', '4', '5', '6', '7', '8', '9' 。每个拨轮可以自由旋转:例如把 '9' 变为 '0','0' 变为 '9' 。每次旋转都只能旋转一个拨轮的一位数字。锁的初始数字为 '0000' ,一个代表四个拨轮的数字的字符串。列表 deadends 包含了一组死亡数字,一旦拨轮的数字和列表里的任何一个元素相同,这个锁将会被永久锁定,无法再被旋转。字符串 target 代表可以解锁的数字,你需要给出解锁需要的最小旋转次数,如果无论如何不能解锁,返回 -1 。leetcode: 725 打开转盘锁input1: deadends = ["0201","0101","0102","1212","2002"], target = "0202"output1: 6input2:deadends = ["8888"], target = "0009"output2:1思路:1.2.3.'''def plus_one(s, i):new_s = []for ss in s:new_s.append(ss)if new_s[i] == '9':new_s[i] = '0'else:new_s[i] = str(int(new_s[i]) + 1)return "".join(new_s)def minus_one(s, i):new_s = []for ss in s:new_s.append(ss)if new_s[i] == '0':new_s[i] = '9'else:new_s[i] = str(int(new_s[i]) + 1)return "".join(new_s)q = ['0000']visited = set()step = 0while len(q) != 0:size = len(q)for i in range(size):cur = q[0]q.remove(q[0])if cur in deadends:continueif cur == target:return stepfor j in range(4):up = plus_one(cur, j)if up not in visited:q.append(up)visited.add(up)down = minus_one(cur, j)if down not in visited:q.append(down)visited.add(down)step += 1return -1
【773 滑动拼图】
def slidingPuzzle(board):'''在⼀个 2 x 3 的棋盘 board 上有 5 块卡⽚,⽤数字 1~5 来表示 , 以及⼀块空缺⽤ 0 来表示。⼀次「移动」定义为选择 0 与⼀个相邻的数字(上下左右)进⾏交换。进⾏若⼲次移动,使得 board 的结果是 [[1,2,3],[4,5,0]],则谜板被解开。给出⼀个谜板的初始状态,返回最少可以通过多少次移动解开谜板,如果不能通过移动解开谜板,则返回-1。leetcode: 773 滑动拼图input:board = [[1,2,3],[4,0,5]]output:1input:board = [[1,2,3],[5,4,0]]output:-1input:board = [[4,1,2],[5,0,3]]output:5思路:1.2.3.'''m, n, target = 2, 3, "123450"start_str = ""for i in range(m):for j in range(n):# 转换为一维的start_str += str(board[i][j])neighbor = [[1, 3],[0, 4, 2],[1, 5],[0, 4],[3, 1, 5],[4, 2]]def modify_str(s, i, v):if i >= len(s):print("[warning] can't modify string!")return sa = [_ for _ in s]a[i] = vnew_s = "".join(a)return new_sq, visited, step = [start_str], set(start_str), 0while len(q) != 0:size = len(q)for i in range(size):cur = q[0]q.remove(q[0])if target == cur:print(step)return stepidx = 0for i in range(6):if cur[i] == '0':idx = ibreakfor j in neighbor[idx]:before, after = cur[i], cur[j]new_board = modify_str(cur, i, after)new_board = modify_str(new_board, j, before)if new_board not in visited:q.append(new_board)visited.add(new_board)step += 1print(-1)return -1
【测试例】
if __name__ == "__main__":# print(openLock(["0201", "0101", "0102", "1212", "2002"], "0202"))# print(openLock(["8888"], '0009'))slidingPuzzle([[1,2,3],[4,0,5]])slidingPuzzle( [[1,2,3],[5,4,0]])slidingPuzzle([[4,1,2],[5,0,3]])
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