LeetCode 542. 01 Matrix–C++解法–动态规划
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大部分题目C++，Python，Java的解法都有。
题目地址：01 Matrix - LeetCode
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.
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 topfor (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 rightfor (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;}};
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