959. Regions Cut By Slashes

Posted 2021-01-29 lzyerste

https://leetcode.com/problems/regions-cut-by-slashes/

题解1，搜索

BFS搜索能连成一片的区域。

class Solution:
    def regionsBySlashes(self, grid):
        """
        :type grid: List[str]
        :rtype: int
        """
        from collections import deque
        
        N = len(grid)
        G = dict()
        # (r, c, d): row, col, direction
        count = 0
        DIR = "UDLR"  # 上下左右
        # 边界
        for r in [-1, N]:
            for c in range(-1, N + 1):
                for d in DIR:
                    G[(r, c, d)] = count
        
        for c in [-1, N]:
            for r in range(-1, N + 1):
                for d in DIR:
                    G[(r, c, d)] = count
        
        D = {
            ("U", " "): [(-1, 0, "D"), (0, 0, "L"), (0, 0, "R")],
            ("U", "/"): [(-1, 0, "D"), (0, 0, "L")],
            ("U", "\"): [(-1, 0, "D"), (0, 0, "R")],
            ("D", " "): [(1, 0, "U"), (0, 0, "L"), (0, 0, "R")],
            ("D", "/"): [(1, 0, "U"), (0, 0, "R")],
            ("D", "\"): [(1, 0, "U"), (0, 0, "L")],
            ("L", " "): [(0, -1, "R"), (0, 0, "U"), (0, 0, "D")],
            ("L", "/"): [(0, -1, "R"), (0, 0, "U")],
            ("L", "\"): [(0, -1, "R"), (0, 0, "D")],
            ("R", " "): [(0, 1, "L"), (0, 0, "U"), (0, 0, "D")],
            ("R", "/"): [(0, 1, "L"), (0, 0, "D")],
            ("R", "\"): [(0, 1, "L"), (0, 0, "U")],
        }
        
        def handle(p, s):
            r, c, d = p
            for dr, dc, dd in D[(d, s)]:
                yield (r + dr, c + dc, dd)
        
        for r in range(N):
            for c in range(N):
                for d in DIR:
                    p = (r, c, d)
                    if p in G:
                        continue  # 已访问
                    
                    count += 1
                    Q = deque([p])
                    while Q:
                        e = Q.popleft()
                        if e in G:
                            continue
                        r1, c1, d1 = e
                        G[e] = count
                        for neighbor in handle(e, grid[r1][c1]):
                            Q.append(neighbor)
        return count

Runtime: 236 ms, faster than 76.69% of Python3 online submissions for Regions Cut By Slashes.