LeetCode – 01 Matrix

题目：

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:

1. The number of elements of the given matrix will not exceed 10,000.
2. There are at least one 0 in the given matrix.
3. The cells are adjacent in only four directions: up, down, left and right.

解题：

1.  BFS:

•   遍历矩阵将所有值为0的位置放入队列， 对于值1 的点设置为INT_MAX（未遍历标识）
•    从queue中取出数字， 遍历周围四个点，越界或者小于当前值+1不更新，由于当前值不会大于INT_MAX（10000个），所有已遍历标识（INT_MAX）都大于当前值，需要设置新值。
• 设置完毕新值后加入队列，继续向周围扩散判断。

实现：

``````class Solution {
public:
vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector<vector<int>> dirs{{0,-1},{-1,0},{0,1},{1,0}};
queue<pair<int, int>> q;

for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
if (matrix[i][j] == 0){
q.push({i, j});
}
else{
matrix[i][j] = INT_MAX;
}
}
}

while(!q.empty()) {
pair<int, int> t = q.front();
q.pop();
for (auto dir: dirs){
int x = t.first + dir[0];
int y = t.second + dir[1];
if (x<0 || x>=m || y<0 || y>=n ||\
matrix[x][y] <= matrix[t.first][t.second] +1){

continue;
}
else {
matrix[x][y] = matrix[t.first][t.second] + 1;
q.push({x,y});
}
}
}
return matrix;
}
};``````

2. 动态规划

• 初始化非0值未INT_MAX -1标识未达到
• 两次遍历左，上方向：取其中较小值+1，  第二次遍历右，下方向：排除0，1值后，取其中较小值+1

实现：

``````class Solution {
public:
vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
int m = matrix.size();
int n = matrix[0].size();
vector<vector<int>> res(m, vector<int>(n, INT_MAX-1));

for (int i=0; i<m; i++){
for (int j=0; j<n; j++){
if (matrix[i][j] == 0){
res[i][j] = 0;
}
else{
if (i > 0){
res[i][j] = min(res[i][j], res[i-1][j]+1);
}
if (j > 0){
res[i][j] = min(res[i][j], res[i][j-1]+1);
}
}
}
}

for (int i=m-1 ;i >=0 ; i--){
for (int j= n-1; j>=0; j--){
if (res[i][j]!=0 && res[i][j]!=1){
if (i < m-1) {
res[i][j] =  min(res[i][j], res[i+1][j]+1);
}
if (j < n-1) {
res[i][j] =  min(res[i][j], res[i][j+1]+1);
}
}
}
}

return res;
}
};
``````

[LeetCode] 01 Matrix 零一矩阵