第十五周：( LeetCode542) 01 Matrix（c++）

原题：
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.

思路：题目的意思是找出矩阵中非0的点和最近的0点的距离，已知上、下、左、右四个方向相邻的点的距离为1。其实就是逐个点做个广度优先搜索。

代码：

class Solution {public:    int bfs(int x,int y,vector<vector<int>>& matrix,int m,int n){        int distance=0;        queue<pair<pair<int,int>,int>> q;        q.push(make_pair(make_pair(x,y),distance));        while(!q.empty()){            x=q.front().first.first;            y=q.front().first.second;            distance=q.front().second;            if(matrix[x][y]==0)                return distance;            if(x-1>=0) q.push(make_pair(make_pair(x-1,y),distance+1));            if(y-1>=0) q.push(make_pair(make_pair(x,y-1),distance+1));            if(x+1<m) q.push(make_pair(make_pair(x+1,y),distance+1));            if(y+1<n) q.push(make_pair(make_pair(x,y+1),distance+1));            q.pop();        }    }    vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {        int m=matrix.size();        if(m==0)  return matrix;        int n=matrix[0].size();        for(int i=0;i<m;i++)            for(int j=0;j<n;j++)                if(matrix[i][j]!=0)                    matrix[i][j]=bfs(i,j,matrix,m,n);        return matrix;    }};