LeetCode:01 Matrix

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

思路：直接深度优先遍历即可，注意记录下当前最短路径，有可能超时，以及对于返回值的处理，需要判断是否是最大INT32，否则+1会溢出。

int getNearset(vector<vector<int>>& matrix, int i, int j,int m,int n,vector<vector<int>>&path,int dep,int mindep)
{
    if (i < 0 || i >= m || j < 0 || j >= n || matrix[i][j]==2 || dep-2>mindep)
        return 0x7fffffff;
    if (path[i][j] != -1)
        return path[i][j];
    if (matrix[i][j] == 0)
        return path[i][j]=0;
    matrix[i][j] = 2;
    int a = getNearset(matrix, i - 1, j, m, n, path,dep+1,0x7fffffff);
    int b = getNearset(matrix, i + 1, j, m, n, path,dep+1,a);
    int c = getNearset(matrix, i, j - 1, m, n, path,dep+1,min(a,b));
    int d = getNearset(matrix, i, j + 1, m, n, path,dep+1,min(min(a,b),c));
    int mindis = min(min(a,b), min(c,d));
    matrix[i][j] = 1;
    return mindis==0x7fffffff?0x7fffffff:mindis+1;
}
vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) 
{
    vector<vector<int>>r;
    int m = matrix.size();
    if (m == 0)
        return r;
    int n = matrix[0].size();
    vector<vector<int>>path(m, vector<int>(n, -1));
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            path[i][j] = getNearset(matrix, i, j, m, n, path,0,0x7fffffff);
    return path;
}

 如果你有任何疑问或新的想法，欢迎在下方评论。