{"id":7290,"date":"2020-08-23T01:46:10","date_gmt":"2020-08-23T08:46:10","guid":{"rendered":"https:\/\/zxi.mytechroad.com\/blog\/?p=7290"},"modified":"2020-08-23T01:47:00","modified_gmt":"2020-08-23T08:47:00","slug":"leetcode-1559-detect-cycles-in-2d-grid","status":"publish","type":"post","link":"https:\/\/zxi.mytechroad.com\/blog\/graph\/leetcode-1559-detect-cycles-in-2d-grid\/","title":{"rendered":"\u82b1\u82b1\u9171 LeetCode 1559. Detect Cycles in 2D Grid"},"content":{"rendered":"\n

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

A cycle is a path of length 4 or more<\/strong> 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<\/strong> of the current cell.<\/p>\n\n\n\n

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

Return true<\/code> if any cycle of the same value exists in grid<\/code>, otherwise, return false<\/code>.<\/p>\n\n\n\n

Example 1:<\/strong><\/p>\n\n\n\n

\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[\"a\",\"a\",\"a\",\"a\"],[\"a\",\"b\",\"b\",\"a\"],[\"a\",\"b\",\"b\",\"a\"],[\"a\",\"a\",\"a\",\"a\"]]\nOutput:<\/strong> true\nExplanation: <\/strong>There are two valid cycles shown in different colors in the image below:\n\n<\/pre>\n\n\n\n
Example 2:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[\"c\",\"c\",\"c\",\"a\"],[\"c\",\"d\",\"c\",\"c\"],[\"c\",\"c\",\"e\",\"c\"],[\"f\",\"c\",\"c\",\"c\"]]\nOutput:<\/strong> true\nExplanation: <\/strong>There is only one valid cycle highlighted in the image below:\n\n<\/pre>\n\n\n\n
Example 3:<\/strong><\/p>\n\n\n\n
\"\"\/<\/figure>\n\n\n\n
Input:<\/strong> grid = [[\"a\",\"b\",\"b\"],[\"b\",\"z\",\"b\"],[\"b\",\"b\",\"a\"]]\nOutput:<\/strong> false\n<\/pre>\n\n\n\n
Constraints:<\/strong><\/p>\n\n\n\n
  • m == grid.length<\/code><\/li>
  • n == grid[i].length<\/code><\/li>
  • 1 <= m <= 500<\/code><\/li>
  • 1 <= n <= 500<\/code><\/li>
  • grid<\/code> consists only of lowercase English letters.<\/li><\/ul>\n\n\n\n
    Solution: DFS<\/strong><\/h2>\n\n\n\n
    Finding a cycle in an undirected graph => visiting a node that has already been visited and it’s not the parent node of the current node.
    b b
    b b
    null -> (0, 0) -> (0, 1) -> (1, 1) -> (1, 0) -> (0, 0)
    The second time we visit (0, 0) which has already been visited before and it’s not the parent of the current node (1, 0) ( (1, 0)’s parent is (1, 1) ) which means we found a cycle.<\/p>\n\n\n\n
    Time complexity: O(m*n)
    Space complexity: O(m*n)<\/p>\n\n\n\n
    \n
    C++<\/h2>\n
    \n\n
    \nclass Solution {\npublic:\n  bool containsCycle(vector>& grid) {\n    const int m = grid.size();\n    const int n = grid[0].size();\n    vector> seen(m, vector(n));\n    vector dirs{0, 1, 0, -1, 0};\n    function dfs = [&](int i, int j, int pi, int pj) {\n      ++seen[i][j];\n      for (int d = 0; d < 4; ++d) {\n        int ni = i + dirs[d];\n        int nj = j + dirs[d + 1];\n        if (ni < 0 || nj < 0 || ni >= m || nj >= n) continue;\n        if (grid[ni][nj] != grid[i][j]) continue;\n        if (!seen[ni][nj]) {\n          if (dfs(ni, nj, i, j)) return true;\n        } else if (ni != pi || nj != pj) {\n          return true;\n        }       \n      }\n      return false;\n    };    \n    for (int i = 0; i < m; ++i)\n      for (int j = 0; j < n; ++j)\n        if (!seen[i][j]++ && dfs(i, j, -1, -1)) \n          return true;\n    return false;\n  }\n};\n<\/pre>\n<\/div><\/div>\n","protected":false},"excerpt":{"rendered":"
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