在给定的网格中，每个像元可以具有以下三个值之一：
值0表示一个空单元格；
值1代表鲜橙色；
值2代表烂橙。
每分钟，与腐烂的橙子（4方向）相邻的任何新鲜的橙子都会变烂。
返回在没有任何单元格显示鲜橙色之前必须经过的最小分钟数。如果这不可能，则返回-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
预期：4
最佳答案
编辑1
您计算分钟数的方法是错误的，每当一个烂橘子至少将一个新鲜的橘子变成一个烂橘子时，就增加分钟数。因此，每分钟的结果分钟数还取决于您对烂橘子进行迭代的顺序，这是错误的。
橘子必须平行变烂，迭代到网格中的顺序必须不相关。
如果我在您的程序中添加每分钟的网格打印结果，则会得到：
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
从您的建议中更正的方式可以是：
#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;
}
编译与执行：
/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，这种方法比下面的方法更有效，因为每个烂橙只考虑一次
所需时间取决于以下事实：是否考虑到对角线在烂橙周围也使新鲜的橙烂。
在我的实现中，我使用预处理器变量DIAG来考虑或不考虑对角线，并使用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