在给定的网格中，每个单元格可以具有以下三个值之一：
值0表示空单元格;值1表示鲜橙色;值2表示烂橙色。每一分钟，与腐烂的橙色相邻（4方向）的任何新鲜橙色都会腐烂。
返回必须经过的最小分钟数，直到没有细胞有新鲜橙色。如果这是不可能的，请返回-1。
class Solution {
public:
int orangesRotting(vector<vector<int>>& grid) {
int m,n;
m = grid.size();
n = grid[0].size();
int i, j, min = 0,flag=0,fresh=0;
int r[4] = {-1,1,0,0};
int c[4] = {0,0,-1,1};
for(i=0;i<m;i++) {
for(j=0;j<n;j++) {
if(grid[i][j]==1)
fresh++;
}
}
queue< pair<int, int> >q;
for(i=0;i<m;i++) {
for(j=0;j<n;j++) {
if(grid[i][j] == 2) {
q.push(make_pair(i,j));
flag=1;
break;
}
}
if(flag==1)
break;
}
while(!q.empty()) {
pair<int,int> p = q.front();
int a = p.first;
int b = p.second;
int x=0;
q.pop();
for(i=0;i<4;i++) {
for(j=0;j<4;j++) {
int rr = a + r[i];
int cc = b + c[j];
if(rr<0 || cc<0 || rr>=m || cc>=n || grid[rr][cc]==0 || grid[rr][cc] ==2) {
continue;
}
else if(grid[rr][cc] ==1) {
grid[rr][cc] =2;
q.push(make_pair(rr,cc));
fresh--;
x++;
}
}
}
if(x>0) min++;
}
return fresh >0 ? -1:min;
}
};
输入：[[2,1,1]，[1,1,0]，[0,1,1]]
输出：3
预计：4
编辑1
你计算分钟数的方法是错误的，每当一个烂橙色至少变成一个新鲜的橙色变成腐烂的橙色时，你会增加分钟数。因此，每分钟的结果分钟也取决于你在腐烂的橙色上迭代的顺序，这是错误的。
橙子必须平行腐烂，你迭代进入网格的顺序一定不相关。
如果我在程序中添加每分钟打印的网格，它会给出：
t = 0
211
110
011
t = 1
221
220
011
t = 2
221
220
021
t = 3
222
220
022
t = 3
222
220
022
t = 3
222
220
022
t = 3
222
220
022
与我的案例比较
编辑2
从您的提案中更正的方法可以是：
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
int orangesRotting(vector<vector<int>>& grid) {
int m,n;
m = grid.size();
n = grid[0].size();
int i, j, min = 0,flag=0,fresh=0;
int r[4] = {-1,1,0,0};
int c[4] = {0,0,-1,1};
queue< pair<int, int>>q;
for(i=0;i<m;i++) {
for(j=0;j<n;j++) {
if (grid[i][j] == 1)
fresh++;
else if (grid[i][j] == 2)
q.push(make_pair(i,j));
}
}
if (fresh == 0)
return 0;
if (q.empty())
return -1;
for (;;) {
#ifdef DEBUG
cout << "t = " << min << endl;
for(i=0;i<m;i++) {
for(j=0;j<n;j++)
cout << grid[i][j];
cout << endl;
}
cout << endl;
#endif
queue< pair<int, int>>qnext;
while (!q.empty()) {
pair<int,int> p = q.front();
int a = p.first;
int b = p.second;
q.pop();
for(i=0;i<4;i++) {
for(j=0;j<4;j++) {
int rr = a + r[i];
int cc = b + c[j];
if (!(rr<0 || cc<0 || rr>=m || cc>=n || grid[rr][cc]==0 || grid[rr][cc] ==2)
&& (grid[rr][cc] ==1)) {
grid[rr][cc] = 2;
qnext.push(make_pair(rr,cc));
fresh--;
}
}
}
}
min += 1;
if (fresh == 0) {
#ifdef DEBUG
cout << "t = " << min << endl;
for(i=0;i<m;i++) {
for(j=0;j<n;j++)
cout << grid[i][j];
cout << endl;
}
cout << endl;
#endif
return min;
}
if (qnext.empty())
return -1;
q = qnext;
}
}
int main()
{
vector<vector<int> > grid;
grid.resize(3);
grid[0].push_back(2);
grid[0].push_back(1);
grid[0].push_back(1);
grid[1].push_back(1);
grid[1].push_back(1);
grid[1].push_back(0);
grid[2].push_back(0);
grid[2].push_back(1);
grid[2].push_back(1);
cout << orangesRotting(grid) << endl;
}
编译和执行：
/tmp % g++ -DDEBUG oo.cc
/tmp % ./a.out
t = 0
211
110
011
t = 1
221
220
011
t = 2
222
220
022
2
请注意，这种方式比下面的方法更有效，因为您只考虑一次每个烂橙色
所需的时间取决于对角线是否被考虑在腐烂的橙色周围以使鲜橙色腐烂的事实。
在这里我的实现，我使用预处理器变量DIAG考虑或不考虑对角线，并且DEBUG每分钟打印或不打印网格：
#include <iostream>
#include <vector>
using namespace std;
enum State { Empty, Fresh, Rotten };
// I do not see the interest of the class so I removed it
// I do not want to modify the input vector so I get it by value
int orangesRotting(vector<vector<State>> grid)
{
int nmins = 0;
const size_t height = grid.size();
if (height == 0)
return -1;
const size_t width = grid[0].size(); // suppose same size for all sub vectors
if (width == 0)
return -1;
// the grid for the next min, do not work on the
// current grid to not see the cells becoming rotten
// in the current step, changes are done simultaneously
vector<vector<State>> nextGrid = grid;
for (;;) {
#ifdef DEBUG
cout << "t = " << nmins << endl;
#endif
bool modified = false;
int nWasFresh = 0;
for (size_t i = 0; i != height; ++i) {
vector<State> & v = grid[i];
for (size_t j = 0; j != width; ++j) {
#ifdef DEBUG
cout << v[j];
#endif
switch (v[j]) {
case Rotten:
{
// make fresh cells around rotten
#ifdef DIAG
const size_t maxh = min(i + 1, height - 1);
const size_t minw = (j == 0) ? j : j - 1;
const size_t maxw = min(j + 1, width - 1);
for (size_t a = (i == 0) ? i : i - 1; a <= maxh; ++a) {
vector<State> & v = grid[a];
for (size_t b = minw; b <= maxw; ++b) {
if (v[b] == Fresh) {
modified = true;
nextGrid[a][b] = Rotten;
}
}
}
#else
if ((i != 0) && (grid[i-1][j] == Fresh)) {
modified = true;
nextGrid[i-1][j] = Rotten;
}
if ((i != (height-1)) && (grid[i+1][j] == Fresh)) {
modified = true;
nextGrid[i+1][j] = Rotten;
}
if ((j != 0) && (grid[i][j-1] == Fresh)) {
modified = true;
nextGrid[i][j-1] = Rotten;
}
if ((j != (width-1)) && (grid[i][j+1] == Fresh)) {
modified = true;
nextGrid[i][j+1] = Rotten;
}
#endif
}
break;
case Fresh:
nWasFresh += 1;
break;
default:
break;
}
}
#ifdef DEBUG
cout << endl;
#endif
}
#ifdef DEBUG
cout << endl;
#endif
if (nWasFresh == 0)
return nmins;
if (!modified)
return -1;
// update grid and time
grid = nextGrid;
nmins += 1;
}
}
int main()
{
vector<vector<State>> grid;
grid.resize(3);
grid[0].push_back(Rotten);
grid[0].push_back(Fresh);
grid[0].push_back(Fresh);
grid[1].push_back(Fresh);
grid[1].push_back(Fresh);
grid[1].push_back(Empty);
grid[2].push_back(Empty);
grid[2].push_back(Fresh);
grid[2].push_back(Fresh);
cout << orangesRotting(grid) << endl;
}
考虑到对角线的编译和执行：
pi@raspberrypi:/tmp $ g++ -DDEBUG -DDIAG -pedantic -Wextra -Wall o.cc
pi@raspberrypi:/tmp $ ./a.out
t = 0
211
110
011
t = 1
221
220
011
t = 2
222
220
022
2
编译和执行时不考虑对角线：
pi@raspberrypi:/tmp $ g++ -DDEBUG -pedantic -Wextra -Wall o.cc
pi@raspberrypi:/tmp $ ./a.out
t = 0
211
110
011
t = 1
221
210
011
t = 2
222
220
011
t = 3
222
220
021
t = 4
222
220
022
4