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.

大致题意

找到矩阵中每个1最近的0的距离

题解

题目类型：BFS、DFS

这道题可以采用每到一个非0节点就深搜一次，比对所有深搜结果，取路径最短的做法，但其中浪费了不少的子问题，有些节点可以利用其他节点计算的结果求出距离。

这里采用BFS，每层BFS表示的距离+1，可以一次完整搜索解决整个距离问题。

过程：将所有0节点加入队列，开始BFS，每层BFS+1的距离，向四周查找（查找过的不再查找）

代码如下

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;
    }
};