Question

1559. 二维网格图中探测环

English Version

题目描述

给你一个二维字符网格数组 grid ，大小为 m x n ，你需要检查 grid 中是否存在 相同值 形成的环。

一个环是一条开始和结束于同一个格子的长度 大于等于 4 的路径。对于一个给定的格子，你可以移动到它上、下、左、右四个方向相邻的格子之一，可以移动的前提是这两个格子有 相同的值 。

同时，你也不能回到上一次移动时所在的格子。比方说，环  (1, 1) -> (1, 2) -> (1, 1) 是不合法的，因为从 (1, 2) 移动到 (1, 1) 回到了上一次移动时的格子。

如果 grid 中有相同值形成的环，请你返回 true ，否则返回 false 。

 

示例 1：

输入：grid = [["a","a","a","a"],["a","b","b","a"],["a","b","b","a"],["a","a","a","a"]]
输出：true
解释：如下图所示，有 2 个用不同颜色标出来的环：

示例 2：

输入：grid = [["c","c","c","a"],["c","d","c","c"],["c","c","e","c"],["f","c","c","c"]]
输出：true
解释：如下图所示，只有高亮所示的一个合法环：

示例 3：

输入：grid = [["a","b","b"],["b","z","b"],["b","b","a"]]
输出：false

 

提示：

• m == grid.length
• n == grid[i].length
• 1 <= m <= 500
• 1 <= n <= 500
• grid 只包含小写英文字母。

解法

方法一

class Solution:
  def containsCycle(self, grid: List[List[str]]) -> bool:
    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    m, n = len(grid), len(grid[0])
    p = list(range(m * n))
    for i in range(m):
      for j in range(n):
        for a, b in [[0, 1], [1, 0]]:
          x, y = i + a, j + b
          if x < m and y < n and grid[x][y] == grid[i][j]:
            if find(x * n + y) == find(i * n + j):
              return True
            p[find(x * n + y)] = find(i * n + j)
    return False

class Solution {
  private int[] p;

  public boolean containsCycle(char[][] grid) {
    int m = grid.length;
    int n = grid[0].length;
    p = new int[m * n];
    for (int i = 0; i < p.length; ++i) {
      p[i] = i;
    }
    int[] dirs = {0, 1, 0};
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        for (int k = 0; k < 2; ++k) {
          int x = i + dirs[k];
          int y = j + dirs[k + 1];
          if (x < m && y < n && grid[i][j] == grid[x][y]) {
            if (find(x * n + y) == find(i * n + j)) {
              return true;
            }
            p[find(x * n + y)] = find(i * n + j);
          }
        }
      }
    }
    return false;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;

  bool containsCycle(vector<vector<char>>& grid) {
    int m = grid.size(), n = grid[0].size();
    p.resize(m * n);
    for (int i = 0; i < p.size(); ++i) p[i] = i;
    vector<int> dirs = {0, 1, 0};
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        for (int k = 0; k < 2; ++k) {
          int x = i + dirs[k], y = j + dirs[k + 1];
          if (x < m && y < n && grid[x][y] == grid[i][j]) {
            if (find(x * n + y) == find(i * n + j)) return 1;
            p[find(x * n + y)] = find(i * n + j);
          }
        }
      }
    }
    return 0;
  }

  int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
  }
};

func containsCycle(grid [][]byte) bool {
  m, n := len(grid), len(grid[0])
  p := make([]int, m*n)
  for i := range p {
    p[i] = i
  }
  var find func(x int) int
  find = func(x int) int {
    if p[x] != x {
      p[x] = find(p[x])
    }
    return p[x]
  }
  dirs := []int{1, 0, 1}
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      for k := 0; k < 2; k++ {
        x, y := i+dirs[k], j+dirs[k+1]
        if x < m && y < n && grid[x][y] == grid[i][j] {
          if find(x*n+y) == find(i*n+j) {
            return true
          }
          p[find(x*n+y)] = find(i*n + j)
        }
      }
    }
  }
  return false
}

impl Solution {
  #[allow(dead_code)]
  pub fn contains_cycle(grid: Vec<Vec<char>>) -> bool {
    let n = grid.len();
    let m = grid[0].len();
    let mut d_set: Vec<usize> = vec![0; n * m];

    // Initialize the disjoint set
    for i in 0..n * m {
      d_set[i] = i;
    }

    // Traverse the grid
    for i in 0..n {
      for j in 0..m {
        if i + 1 < n && grid[i + 1][j] == grid[i][j] {
          // Check the below cell
          let p_curr = Self::find(i * m + j, &mut d_set);
          let p_below = Self::find((i + 1) * m + j, &mut d_set);
          if p_curr == p_below {
            return true;
          }
          // Otherwise, union the two cells
          Self::union(p_curr, p_below, &mut d_set);
        }
        // Same to the right cell
        if j + 1 < m && grid[i][j + 1] == grid[i][j] {
          let p_curr = Self::find(i * m + j, &mut d_set);
          let p_right = Self::find(i * m + (j + 1), &mut d_set);
          if p_curr == p_right {
            return true;
          }
          // Otherwise, union the two cells
          Self::union(p_curr, p_right, &mut d_set);
        }
      }
    }

    false
  }

  #[allow(dead_code)]
  fn find(x: usize, d_set: &mut Vec<usize>) -> usize {
    if d_set[x] != x {
      d_set[x] = Self::find(d_set[x], d_set);
    }
    d_set[x]
  }

  #[allow(dead_code)]
  fn union(x: usize, y: usize, d_set: &mut Vec<usize>) {
    let p_x = Self::find(x, d_set);
    let p_y = Self::find(y, d_set);
    d_set[p_x] = p_y;
  }
}

/**
 * @param {character[][]} grid
 * @return {boolean}
 */
var containsCycle = function (grid) {
  const m = grid.length;
  const n = grid[0].length;
  let p = Array.from({ length: m * n }, (_, i) => i);
  function find(x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
  const dirs = [0, 1, 0];
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      for (let k = 0; k < 2; ++k) {
        const x = i + dirs[k];
        const y = j + dirs[k + 1];
        if (x < m && y < n && grid[x][y] == grid[i][j]) {
          if (find(x * n + y) == find(i * n + j)) {
            return true;
          }
          p[find(x * n + y)] = find(i * n + j);
        }
      }
    }
  }
  return false;
};