LeetCode 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.

大致題意

找到矩陣中每個1最近的0的距離

題解

題目型別：BFS、DFS

這道題可以採用每到一個非0節點就深搜一次，比對所有深搜結果，取路徑最短的做法，但其中浪費了不少的子問題，有些節點可以利用其他節點計算的結果求出距離。

這裡採用BFS，每層BFS表示的距離+1，可以一次完整搜尋解決整個距離問題。

過程：將所有0節點加入佇列，開始BFS，每層BFS+1的距離，向四周查詢（查詢過的不再查詢）

程式碼如下

class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& A) {
        vector<vector<int> > ans;
        queue<pair<int, int> > Q;
        const int D[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
        int n, m, row, col, i, j;
        pair<int, int> u;
        n = A.size();
        m = A[0].size();
        ans.resize(n);
        for (i=0; i<n; i++) ans[i].resize(m);
        for (i=0; i<n; i++) 
            for (j=0; j<m; j++)
                if (A[i][j] == 0) {
                    Q.push(make_pair(i, j));
                    ans[i][j] = 0;
                }
                else ans[i][j] = -1;
        while (!Q.empty()) {
            u = Q.front();
            Q.pop();
            for (i=0; i<4; i++) {
                row = u.first + D[i][0];
                col = u.second + D[i][1];
                if (row >= 0 && row < n && col >= 0 && col < m && ans[row][col] == -1) {
                    ans[row][col] = ans[u.first][u.second] + 1;
                    Q.push(make_pair(row, col));
                }
            }
        }        
        return ans;
    }
};