LeetCode 542. 01 Matrix--C++解法--动态规划

LeetCode 542. 01 Matrix–C++解法–动态规划

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LeetCode题解专栏：LeetCode题解
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大部分题目C++，Python，Java的解法都有。

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题目地址：01 Matrix - LeetCode

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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.

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这道题目乍一看应该用动态规划做，因为2个样例都可以用动态规划解决。但看了提示后，发现简单的动态规划不行，因为有4个方向。

然后仔细考虑后发现必须要动态规划，不然这道题不能解，时间复杂度太高了。

想找到离1最近0，这个0只能在1的4个方向：左上，左下，右下，右上。

那么让程序从左上遍历一遍，再从右下遍历一遍，即可得到解。

C++解法如下：

class Solution {
public:
    vector<vector<int> > updateMatrix(vector<vector<int> > &matrix) {
        int rows = matrix.size();
        if (rows == 0)
            return matrix;
        int cols = matrix[0].size();
        vector<vector<int> > dist(rows, vector<int>(cols, INT_MAX - 100000));
        //First pass: check for left and top
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (matrix[i][j] == 0)
                    dist[i][j] = 0;
                else {
                    if (i > 0)
                        dist[i][j] = min(dist[i][j], dist[i - 1][j] + 1);
                    if (j > 0)
                        dist[i][j] = min(dist[i][j], dist[i][j - 1] + 1);
                }
            }
        }

        //Second pass: check for bottom and right
        for (int i = rows - 1; i >= 0; i--) {
            for (int j = cols - 1; j >= 0; j--) {
                if (i < rows - 1)
                    dist[i][j] = min(dist[i][j], dist[i + 1][j] + 1);
                if (j < cols - 1)
                    dist[i][j] = min(dist[i][j], dist[i][j + 1] + 1);
            }
        }
        return dist;
    }
};