【LeetCode 542】 01 Matrix

題目描述

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.

思路

bfs，從零步的位置開始拓展。

代碼

class Solution {
public:
    vector<int> dirs {1, 0, -1, 0, 1};
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int n = matrix.size(), m = matrix[0].size();
        vector<vector<int>> res(n, vector<int>(m, INT_MAX));
        queue<pair<int, int>> que;
        
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                if (matrix[i][j] == 0) {
                    res[i][j] = 0;
                    que.push({i, j});
                }
            }
        }
        
        int step = 1;
        while(!que.empty()) {
            int size = que.size();
            while(size--) {
                pair<int, int> tmp = que.front();
                que.pop();
                int x = tmp.first;
                int y = tmp.second;
                for (int i=0; i<4; ++i) {
                    int nx = x + dirs[i];
                    int ny = y + dirs[i+1];
                    if (nx <0 || nx >= n || ny < 0 || ny >= m) continue;
                    if (res[nx][ny] <= res[x][y] + 1) continue;
                    res[nx][ny] = res[x][y] + 1;
                    que.push({nx, ny});
                }
            }
        }
        
        return res;
    }
};