Question

542. 01 Matrix

中文文档

Description

Given an m x n binary matrix mat, return _the distance of the nearest _0_ for each cell_.

The distance between two adjacent cells is 1.

 

Example 1:

Input: mat = [[0,0,0],[0,1,0],[0,0,0]]
Output: [[0,0,0],[0,1,0],[0,0,0]]

Example 2:

Input: mat = [[0,0,0],[0,1,0],[1,1,1]]
Output: [[0,0,0],[0,1,0],[1,2,1]]

 

Constraints:

• m == mat.length
• n == mat[i].length
• 1 <= m, n <= 104
• 1 <= m * n <= 104
• mat[i][j] is either 0 or 1.
• There is at least one 0 in mat.

Solutions

Solution 1

class Solution:
  def updateMatrix(self, mat: List[List[int]]) -> List[List[int]]:
    m, n = len(mat), len(mat[0])
    ans = [[-1] * n for _ in range(m)]
    q = deque()
    for i, row in enumerate(mat):
      for j, x in enumerate(row):
        if x == 0:
          ans[i][j] = 0
          q.append((i, j))
    dirs = (-1, 0, 1, 0, -1)
    while q:
      i, j = q.popleft()
      for a, b in pairwise(dirs):
        x, y = i + a, j + b
        if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:
          ans[x][y] = ans[i][j] + 1
          q.append((x, y))
    return ans

class Solution {
  public int[][] updateMatrix(int[][] mat) {
    int m = mat.length, n = mat[0].length;
    int[][] ans = new int[m][n];
    for (int[] row : ans) {
      Arrays.fill(row, -1);
    }
    Deque<int[]> q = new ArrayDeque<>();
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (mat[i][j] == 0) {
          q.offer(new int[] {i, j});
          ans[i][j] = 0;
        }
      }
    }
    int[] dirs = {-1, 0, 1, 0, -1};
    while (!q.isEmpty()) {
      int[] p = q.poll();
      int i = p[0], j = p[1];
      for (int k = 0; k < 4; ++k) {
        int x = i + dirs[k], y = j + dirs[k + 1];
        if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
          ans[x][y] = ans[i][j] + 1;
          q.offer(new int[] {x, y});
        }
      }
    }
    return ans;
  }
}

class Solution {
public:
  vector<vector<int>> updateMatrix(vector<vector<int>>& mat) {
    int m = mat.size(), n = mat[0].size();
    vector<vector<int>> ans(m, vector<int>(n, -1));
    queue<pair<int, int>> q;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        if (mat[i][j] == 0) {
          ans[i][j] = 0;
          q.emplace(i, j);
        }
      }
    }
    vector<int> dirs = {-1, 0, 1, 0, -1};
    while (!q.empty()) {
      auto p = q.front();
      q.pop();
      for (int i = 0; i < 4; ++i) {
        int x = p.first + dirs[i];
        int y = p.second + dirs[i + 1];
        if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1) {
          ans[x][y] = ans[p.first][p.second] + 1;
          q.emplace(x, y);
        }
      }
    }
    return ans;
  }
};

func updateMatrix(mat [][]int) [][]int {
  m, n := len(mat), len(mat[0])
  ans := make([][]int, m)
  for i := range ans {
    ans[i] = make([]int, n)
    for j := range ans[i] {
      ans[i][j] = -1
    }
  }
  type pair struct{ x, y int }
  var q []pair
  for i, row := range mat {
    for j, v := range row {
      if v == 0 {
        ans[i][j] = 0
        q = append(q, pair{i, j})
      }
    }
  }
  dirs := []int{-1, 0, 1, 0, -1}
  for len(q) > 0 {
    p := q[0]
    q = q[1:]
    for i := 0; i < 4; i++ {
      x, y := p.x+dirs[i], p.y+dirs[i+1]
      if x >= 0 && x < m && y >= 0 && y < n && ans[x][y] == -1 {
        ans[x][y] = ans[p.x][p.y] + 1
        q = append(q, pair{x, y})
      }
    }
  }
  return ans
}

function updateMatrix(mat: number[][]): number[][] {
  const [m, n] = [mat.length, mat[0].length];
  const ans: number[][] = Array.from({ length: m }, () => Array.from({ length: n }, () => -1));
  const q: [number, number][] = [];
  for (let i = 0; i < m; ++i) {
    for (let j = 0; j < n; ++j) {
      if (mat[i][j] === 0) {
        q.push([i, j]);
        ans[i][j] = 0;
      }
    }
  }
  const dirs: number[] = [-1, 0, 1, 0, -1];
  while (q.length) {
    const [i, j] = q.shift()!;
    for (let k = 0; k < 4; ++k) {
      const [x, y] = [i + dirs[k], j + dirs[k + 1]];
      if (x >= 0 && x < m && y >= 0 && y < n && ans[x][y] === -1) {
        ans[x][y] = ans[i][j] + 1;
        q.push([x, y]);
      }
    }
  }
  return ans;
}

use std::collections::VecDeque;

impl Solution {
  #[allow(dead_code)]
  pub fn update_matrix(mat: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
    let n: usize = mat.len();
    let m: usize = mat[0].len();
    let mut ret_vec: Vec<Vec<i32>> = vec![vec![-1; m]; n];
    // The inner tuple is of <X, Y, Current Count>
    let mut the_q: VecDeque<(usize, usize)> = VecDeque::new();
    let traverse_vec: Vec<(i32, i32)> = vec![(-1, 0), (1, 0), (0, 1), (0, -1)];

    // Initialize the queue
    for i in 0..n {
      for j in 0..m {
        if mat[i][j] == 0 {
          // For the zero cell, enqueue at first
          the_q.push_back((i, j));
          // Set to 0 in return vector
          ret_vec[i][j] = 0;
        }
      }
    }

    while !the_q.is_empty() {
      let (x, y) = the_q.front().unwrap().clone();
      the_q.pop_front();
      for pair in &traverse_vec {
        let cur_x = pair.0 + (x as i32);
        let cur_y = pair.1 + (y as i32);
        if
          Solution::check_bounds(cur_x, cur_y, n as i32, m as i32) &&
          ret_vec[cur_x as usize][cur_y as usize] == -1
        {
          // The current cell has not be updated yet, and is also in bound
          ret_vec[cur_x as usize][cur_y as usize] = ret_vec[x][y] + 1;
          the_q.push_back((cur_x as usize, cur_y as usize));
        }
      }
    }

    ret_vec
  }

  #[allow(dead_code)]
  pub fn check_bounds(i: i32, j: i32, n: i32, m: i32) -> bool {
    i >= 0 && i < n && j >= 0 && j < m
  }
}