LeetCode 542题解析-优快云博客

本文介绍了解决LeetCode上542题“01 Matrix”的算法思路及实现，该题要求在一个由0和1组成的矩阵中，计算每个元素到最近0的距离。文章详细解释了如何使用广度优先搜索算法进行高效求解。

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LeetCode Algorithm部分.542

【542】01 Matrix

Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
The distance between two adjacent cells is 1.
Example 1:
Input:

0 0 0
0 1 0
0 0 0

Output:

0 0 0
0 1 0
0 0 0

Example 2:
Input:

0 0 0
0 1 0
1 1 1

Output:

0 0 0
0 1 0
1 2 1

Note:
The number of elements of the given matrix will not exceed 10,000.
There are at least one 0 in the given matrix.
The cells are adjacent in only four directions: up, down, left and right.

该题的大意为，存在一个01矩阵，找到每个元素距离最近0元素的距离。

class Solution {
public:
  vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
    int M = matrix.size(), N = matrix[0].size();
    queue<pair<int, int>> q;
    for (int i = 0; i < M; ++i) {
      for (int j = 0; j < N; ++j) {
        if (!matrix[i][j]) q.emplace(i, j);
        else matrix[i][j] = INT_MAX;
      }
    }
    int dirs[][2] = {{ -1, 0 }, { 0, 1 }, { 1, 0 }, { 0, -1 }};
    while (q.size()) {
      auto p = q.front();
      q.pop();
      for (auto dir : dirs) {
        int x = p.first + dir[0], y = p.second + dir[1];
        if (x >= 0 && x < M && y >= 0 && y < N && matrix[x][y] == INT_MAX) {
          matrix[x][y] = matrix[p.first][p.second] + 1;
          q.emplace(x, y);
        }
      }
    }
    return matrix;

  }
};