{
  "type": "code_options",
  "author": "https://github.com/begeekmyfriend/leetcode",
  "source": "solution.md",
  "exercise_id": "485511cc5a7a4c6e8fdb39cd60d07047",
  "keywords": "数组,回溯,矩阵",
  "title": "单词搜索",
  "desc": [
    {
      "content": "\n<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>",
      "language": "markdown"
    }
  ],
  "answer": [
    {
      "content": "",
      "language": "cpp"
    }
  ],
  "prepared": [
    [
      {
        "content": "",
        "language": "cpp"
      }
    ],
    [
      {
        "content": "",
        "language": "cpp"
      }
    ],
    [
      {
        "content": "",
        "language": "cpp"
      }
    ]
  ],
  "template": {
    "content": "#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}",
    "language": "cpp"
  },
  "node_id": "dailycode-b3f62c12a2624288bb79b4210b44e046",
  "license": "csdn.net",
  "created_at": 1637894158,
  "topic_link": "https://bbs.csdn.net/topics/600470927"
}