{
   "question_id": 79,
   "question_title": "单词搜索",
   "difficulty": "中等",
   "question_content": "<p>给定一个 <code>m x n</code> 二维字符网格 <code>board</code> 和一个字符串单词 <code>word</code> 。如果 <code>word</code> 存在于网格中，返回 <code>true</code> ；否则，返回 <code>false</code> 。</p><p>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。</p><p> </p><p><strong>示例 1：</strong></p><img alt=\"\" src=\"https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0079.Word%20Search/images/word2.jpg\" style=\"width: 322px; height: 242px;\" /><pre><strong>输入：</strong>board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"ABCCED\"<strong><br />输出：</strong>true</pre><p><strong>示例 2：</strong></p><img alt=\"\" src=\"https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0079.Word%20Search/images/word-1.jpg\" style=\"width: 322px; height: 242px;\" /><pre><strong>输入：</strong>board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"SEE\"<strong><br />输出：</strong>true</pre><p><strong>示例 3：</strong></p><img alt=\"\" src=\"https://cdn.jsdelivr.net/gh/doocs/leetcode@main/solution/0000-0099/0079.Word%20Search/images/word3.jpg\" style=\"width: 322px; height: 242px;\" /><pre><strong>输入：</strong>board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"ABCB\"<strong><br />输出：</strong>false</pre><p> </p><p><strong>提示：</strong></p><ul>\t<li><code>m == board.length</code></li>\t<li><code>n = board[i].length</code></li>\t<li><code>1 <= m, n <= 6</code></li>\t<li><code>1 <= word.length <= 15</code></li>\t<li><code>board</code> 和 <code>word</code> 仅由大小写英文字母组成</li></ul><p> </p><p><strong>进阶：</strong>你可以使用搜索剪枝的技术来优化解决方案，使其在 <code>board</code> 更大的情况下可以更快解决问题？</p>",
   "topic_link": "https://bbs.csdn.net/topics/600470927",
   "cpp": "#include <stdio.h>\n#include <stdlib.h>\n#include <stdbool.h>\n#include <string.h>\nstatic bool dfs(char *word, char **board, bool *used,\n\t\t\t\tint row, int col, int row_size, int col_size)\n{\n\tif (board[row][col] != *word)\n\t{\n\t\treturn false;\n\t}\n\tused[row * col_size + col] = true;\n\tif (*(word + 1) == '\\0')\n\t{\n\t\treturn true;\n\t}\n\tbool result = false;\n\tif (row > 0 && !used[(row - 1) * col_size + col])\n\t{\n\t\tresult = dfs(word + 1, board, used, row - 1, col, row_size, col_size);\n\t}\n\tif (!result && row < row_size - 1 && !used[(row + 1) * col_size + col])\n\t{\n\t\tresult = dfs(word + 1, board, used, row + 1, col, row_size, col_size);\n\t}\n\tif (!result && col > 0 && !used[row * col_size + col - 1])\n\t{\n\t\tresult = dfs(word + 1, board, used, row, col - 1, row_size, col_size);\n\t}\n\tif (!result && col < col_size - 1 && !used[row * col_size + col + 1])\n\t{\n\t\tresult = dfs(word + 1, board, used, row, col + 1, row_size, col_size);\n\t}\n\tused[row * col_size + col] = false;\n\treturn result;\n}\nstatic bool exist(char **board, int boardRowSize, int boardColSize, char *word)\n{\n\tint i, j;\n\tint len = strlen(word);\n\tif (len > boardRowSize * boardColSize)\n\t{\n\t\treturn false;\n\t}\n\tbool *used = malloc(boardRowSize * boardColSize);\n\tfor (i = 0; i < boardRowSize; i++)\n\t{\n\t\tfor (j = 0; j < boardColSize; j++)\n\t\t{\n\t\t\tmemset(used, false, boardRowSize * boardColSize);\n\t\t\tif (dfs(word, board, used, i, j, boardRowSize, boardColSize))\n\t\t\t{\n\t\t\t\treturn true;\n\t\t\t}\n\t\t}\n\t}\n\treturn false;\n}\nint main(int argc, char **argv)\n{\n\tif (argc < 3)\n\t{\n\t\tfprintf(stderr, \"Usage: ./test word row1 row2...\\n\");\n\t\texit(-1);\n\t}\n\tprintf(\"%s\\n\", exist(argv + 2, argc - 2, strlen(argv[2]), argv[1]) ? \"true\" : \"false\");\n\treturn 0;\n}",
   "java": "class Solution {\n\tpublic boolean exist(char[][] board, String word) {\n\t\tint cl = board.length;\n\t\tint rl = board[0].length;\n\t\tboolean[][] flag = new boolean[cl][rl];\n\t\tfor (int i = 0; i < cl; i++) {\n\t\t\tfor (int j = 0; j < rl; j++) {\n\t\t\t\tif (find(board, word, flag, i, j, 0))\n\t\t\t\t\treturn true;\n\t\t\t}\n\t\t}\n\n\t\treturn false;\n\t}\n\n\tpublic boolean find(char[][] board, String word, boolean[][] flag, int i, int j, int index) {\n\t\tint cl = board.length;\n\t\tint rl = board[0].length;\n\t\tif (word.length() == index)\n\t\t\treturn true;\n\t\tif (i < 0 || i >= cl || j >= rl || j < 0)\n\t\t\treturn false;\n\t\tif (flag[i][j] || word.charAt(index) != board[i][j])\n\t\t\treturn false;\n\t\tflag[i][j] = true;\n\t\tboolean judge = find(board, word, flag, i - 1, j, index + 1) || find(board, word, flag, i + 1, j, index + 1)\n\t\t\t\t|| find(board, word, flag, i, j - 1, index + 1) || find(board, word, flag, i, j + 1, index + 1);\n\t\tflag[i][j] = false;\n\t\treturn judge;\n\n\t}\n}\n",
   "js": "",
   "python": "class Solution(object):\n\tdef exist(self, board, word):\n\t\t\"\"\"\n\t\t:type board: List[List[str]]\n\t\t:type word: str\n\t\t:rtype: bool\n\t\t\"\"\"\n\t\tcheck_board = [[True] * len(board[0]) for _ in range(len(board))]\n\t\tfor i in range(len(board)):\n\t\t\tfor j in range(len(board[0])):\n\t\t\t\tif board[i][j] == word[0] and check_board:\n\t\t\t\t\tcheck_board[i][j] = False\n\t\t\t\t\tres = self.check_exist(check_board, board, word, 1, len(word), i, j)\n\t\t\t\t\tif res:\n\t\t\t\t\t\treturn True\n\t\t\t\t\tcheck_board[i][j] = True\n\t\treturn False\n\tdef check_exist(self, check_board, board, word, index, ls, row, col):\n\t\tif index == ls:\n\t\t\treturn True\n\t\tfor temp in [(0, 1),(0, -1),(1, 0),(-1, 0)]:\n\t\t\tcurr_row = row + temp[0]\n\t\t\tcurr_col = col + temp[1]\n\t\t\tif curr_row >= 0 and curr_row < len(board) and curr_col >= 0 and curr_col < len(board[0]):\n\t\t\t\tif check_board[curr_row][curr_col] and board[curr_row][curr_col] == word[index]:\n\t\t\t\t\tcheck_board[curr_row][curr_col] = False\n\t\t\t\t\tres = self.check_exist(check_board, board, word, index + 1, len(word), curr_row, curr_col)\n\t\t\t\t\tif res:\n\t\t\t\t\t\treturn res\n\t\t\t\t\tcheck_board[curr_row][curr_col] = True\n\t\treturn False\nif __name__ == \"__main__\":\n\ts = Solution()\n\tprint (s.exist(board = [[\"A\",\"B\",\"C\",\"E\"],[\"S\",\"F\",\"C\",\"S\"],[\"A\",\"D\",\"E\",\"E\"]], word = \"ABCCED\"))",
   "status": 1,
   "keywords": "数组,回溯,矩阵",
   "license": {
      "cpp": "https://github.com/begeekmyfriend/leetcode",
      "python": "https://github.com/qiyuangong/leetcode",
      "java": "https://github.com/zhangyu345293721/leetcode"
   },
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      "python": "https://codechina.csdn.net/csdn/csdn-daily-code/-/jupyter/master/data/notebook/leetcode/ipynb/78/78_python.ipynb?type=file",
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