【leetcode】1559. Detect Cycles in 2D Grid

题目如下：

Given a 2D array of characters grid of size m x n, you need to find if there exists any cycle consisting of the same value in grid.

A cycle is a path of length 4 or more in the grid that starts and ends at the same cell. From a given cell, you can move to one of the cells adjacent to it - in one of the four directions (up, down, left, or right), if it has the same value of the current cell.

Also, you cannot move to the cell that you visited in your last move. For example, the cycle (1, 1) -> (1, 2) -> (1, 1) is invalid because from (1, 2) we visited (1, 1) which was the last visited cell.

Return true if any cycle of the same value exists in grid, otherwise, return false.

Example 1:
Input: grid = [["a","a","a","a"],["a","b","b","a"],["a","b","b","a"],["a","a","a","a"]]
Output: true
Explanation: There are two valid cycles shown in different colors in the image below:
Example 2:
Input: grid = [["c","c","c","a"],["c","d","c","c"],["c","c","e","c"],["f","c","c","c"]]
Output: true
Explanation: There is only one valid cycle highlighted in the image below:
Example 3:
Input: grid = [["a","b","b"],["b","z","b"],["b","b","a"]]
Output: false
Constraints:

m == grid.length

n == grid[i].length

1 <= m <= 500

1 <= n <= 500

grid consists only of lowercase English letters.

解题思路：题目不难，利用dfs/bfs的思想遍历一遍就行了，遍历的过程中记录走过的节点，遇到重复就说明有环。有一点要注意，需要同时记录到达当前节点的上一个节点的位置，往下一个节点走的时候不能重复回到上一个节点，以免掉入死循环。

代码如下：

class Solution(object):
    def containsCycle(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: bool
        """
        visit = {}
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if (i,j) in visit:continue
                visit[(i, j)] = 1
                queue = [(None,None,i,j)]
                while len(queue) > 0:
                    lx,ly,x,y = queue.pop(0)
                    direction = [(0,1),(0,-1),(1,0),(-1,0)]
                    for (x1,y1) in direction:
                        if x + x1 >= 0 and x + x1 < len(grid) and y + y1 >= 0 
                                and y + y1 < len(grid[0]) and grid[x][y] == grid[x+x1][y+y1] 
                            and (lx != x + x1 or ly != y + y1):
                            if (x + x1,y+y1) in visit:
                                return True
                            visit[(x+x1,y+y1)] = 1
                            queue.append((x,y,x+x1,y+y1))
        return False