【Leetcode】剑指Offer 12：矩阵中的路径

给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。 例如，在下面的 3×4 的矩阵中包含单词 “ABCCED”（单词中的字母已标出）。 示例 1： 输入：board = [[“A”,“B”,“C”,“E”],[“S”,“F”,“C”,“S”],[“A”,“D”,“E”,“E”]], word = “ABCCED” 输出：true 示例 2： 输入：board = [[“a”,“b”],[“c”,“d”]], word = “abcd” 输出：false 提示： m == board.length n = board[i].length 1 bool: def judgeLeagel(a:tuple, m:int, n:int) -> bool: return a[0] >= 0 and a[0] = 0 and a[1] bool: currentx, currenty = path[-1] alist = [] alist.append((currentx, currenty + 1)) alist.append((currentx, currenty - 1)) alist.append((currentx + 1, currenty)) alist.append((currentx - 1, currenty)) b = [False, False, False, False] for i, a in enumerate(alist): if judgeLeagel(a, m, n) and board[a[0]][a[1]] == word[0] and a not in path: if len(word) == 1: return True else: temp = copy.deepcopy(path) temp.append(a) b[i] = getpath(temp, word[1:], m, n) if b[i] == True: return True else: return False if len(word) == 1: for l in board: if word in l: return True else: return False alphalist = [] for i, a in enumerate(board): for j, n in enumerate(a): if n == word[0]: alphalist.append((i, j)) if alphalist == []: return False else: for a in alphalist: if getpath([a], word[1:], len(board), len(board[0])): return True else: return False

WA的原因是超时，这是由于深拷贝对内存和计算量都有很大的负担。

AC代码

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def judgeLeagel(a:tuple, m:int, n:int) -> bool:
            return a[0] >= 0 and a[0] = 0 and a[1]  bool:
            currentx, currenty = path[-1]
            alist = []
            alist.append((currentx, currenty + 1))
            alist.append((currentx, currenty - 1))
            alist.append((currentx + 1, currenty))
            alist.append((currentx - 1, currenty))
            b = [False, False, False, False]

            for i, a in enumerate(alist):
                if judgeLeagel(a, m, n) and board[a[0]][a[1]] == word[0] and a not in path:
                    if len(word) == 1:
                        return True
                    else:
                        # temp = copy.deepcopy(path)
                        # temp.append(a)
                        path.append(a)
                        b[i] = getpath(path, word[1:], m, n)
                        path.remove(a)
                        if b[i] == True:
                            return True
            else:
                return False
        if len(word) == 1:
            for l in board:
                if word in l:
                    return True
            else:
                return False

        alphalist = []
        for i, a in enumerate(board):
            for j, n in enumerate(a):
                if n == word[0]:
                    alphalist.append((i, j))
        if alphalist == []:
            return False
        else:
            for a in alphalist:
                if getpath([a], word[1:], len(board), len(board[0])):
                    return True
            else:
                return False

利用remove的方法就能够实现上面的需求，笔者由于最开始只想着引用和深浅拷贝的问题，忽略了直接通过remove一样可以达到这样的效果。许久不做DFS的题目对一些简单的技巧都忘光了。好在写题的时候已经意识到了一部分剪枝的问题，最后勉强AC。