[LeetCode] 01 Matrix

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题目描述

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.
输入是一个只包含0 1 的矩阵，要求输出每个元素和矩阵中0最近距离。

C++ 代码实现

动态规划版本，按照从上到下，从左到右的顺序遍历可以考虑到从左边以及上边到达的最短路径情况。然后再按照从下到上，从右到左的顺序遍历可以考虑到从右边以及下边到达的最短路径。因此动态规划需要一次正序遍历，一次逆序遍历。

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class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int row_size = matrix.size();
        if (row_size == 0) return vector<vector<int>>();
        int col_size = matrix[0].size();
        vector<vector<int>> res(row_size,vector<int>(col_size));
        // forward loop        
        for (int i = 0; i < row_size; i++)
        {
            for (int j = 0; j < col_size; j++)
            {
                if (matrix[i][j] == 0)
                {
                    res[i][j] = 0;
                }
                else
                {
                    int left = INT_MAX - 1;
                    int top = INT_MAX - 1;
                    if (i - 1 >= 0) top = res[i - 1][j];
                    if (j - 1 >= 0) left = res[i][j - 1];
                    res[i][j] = min(INT_MAX - 1, min(left,top) + 1);
                }
            }
        }
        // reverse loop
        
        for (int i = row_size - 1; i >= 0; i--)
        {
            for (int j = col_size - 1; j >= 0; j--)
            {
                if (res[i][j] != 0 && res[i][j] != 1)
                {
                    int down = INT_MAX - 1;
                    int right = INT_MAX - 1;
                    if (i + 1 < row_size) down = res[i + 1][j];
                    if (j + 1 < col_size) right = res[i][j + 1];
                    res[i][j] = min(res[i][j], min(down,right) + 1);
                }
            }
        }
        
        return res;
    }
};

BFS 的解法如下

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class Solution {
public:
    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
        int row_size = matrix.size();
        int col_size = matrix[0].size();
        queue<pair<int,int>> queue;
        for (int i = 0; i < row_size; i++)
        {
            for (int j = 0; j < col_size; j++)
            {
                if (matrix[i][j] == 0)
                {
                    queue.push({i, j});
                }
                else
                {
                    matrix[i][j] = INT_MAX;
                }
            }
        }
        vector<pair<int,int>> directions = {{0,1},{1,0},{0,-1},{-1,0}};
        
        while(!queue.empty())
        {
            const auto front = queue.front();
            queue.pop();
            int x = front.first;
            int y = front.second;
            for (const auto direction : directions)
            {
                int newx = x + direction.first;
                int newy = y + direction.second;
                if (   newx < 0 
                    || newy < 0 
                    || newx >= row_size 
                    || newy >= col_size 
                    || matrix[newx][newy] <= matrix[x][y] + 1)
                {
                    continue;
                }
                matrix[newx][newy] = matrix[x][y] + 1;
                queue.push({newx, newy});
            }
        }
        
        return matrix;
    }
};

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