[Leetcode] #542 01 Matrix (BFS)

Discription

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

Solution

class Solution {
public:
	vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
		if (matrix.empty())
			return matrix;
		queue<pair<int, int>> que;
		int rows = matrix.size(), cols = matrix[0].size();
		vector<vector<int>> res(rows, vector<int>(cols, 0));
		vector<vector<int>> dirs = { { -1,0 },{ 1,0 },{ 0,-1 },{ 0,1 } };

		for (int i = 0; i < rows; i++) {
			for (int j = 0; j < cols; j++) {
				if (matrix[i][j] == 0)
					que.push({ i,j });
				else
					res[i][j] = -1;
			}
		}

		while (!que.empty()) {
			pair<int, int> point = que.front();
			int x = point.first, y = point.second;
			que.pop();
			for (int i = 0; i < dirs.size(); i++) {
				int nx = x + dirs[i][0], ny = y + dirs[i][1];
				if (nx >= 0 && nx < rows && ny >= 0 && ny < cols && res[nx][ny] == -1) {
					res[nx][ny] = res[x][y] + 1;
					que.push({ nx,ny });
				}
			}
		}
		return res;
	}
};

GitHub-Leetcode: https://github.com/wenwu313/LeetCode