1.图像渲染
图像渲染
dfs解决代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
int prev;
vector<vector<int>> ret;
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
if(image[sr][sc] == color) return image;
m = image.size(), n = image[0].size();
prev = image[sr][sc];
dfs(image, sr, sc, color);
return image;
}
void dfs(vector<vector<int>>& image, int i, int j, int color)
{
image[i][j] = color;
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && image[x][y] == prev)
{
dfs(image, x, y, color);
}
}
}
};bfs解决代码：
class Solution {
public:
typedef pair<int, int> PII;
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
int prev = image[sr][sc];
if(prev == color) return image;
int m = image.size(), n = image[0].size();
queue<PII> q;
q.push({sr, sc});
while(!q.empty())
{
auto& [a, b] = q.front();
image[a][b] = color;
q.pop();
for(int k = 0; k < 4; k++)
{
int x = a + dx[k], y = b + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && image[x][y] == prev)
{
q.push({x, y});
}
}
}
return image;
}
};2.岛屿数量
岛屿数量
dfs解决代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n, ret;
bool vis[301][301];
int numIslands(vector<vector<char>>& grid) {
m = grid.size(), n = grid[0].size();
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if(!vis[i][j] && grid[i][j] == '1')
{
ret++;
dfs(grid, i, j); //标记联通1
}
}
}
return ret;
}
void dfs(vector<vector<char>>& grid, int i, int j)
{
vis[i][j] = true;
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == '1')
{
dfs(grid, x, y);
}
}
}
};bfs解决代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n, ret;
bool vis[301][301];
int numIslands(vector<vector<char>>& grid) {
m = grid.size(), n = grid[0].size();
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if(!vis[i][j] && grid[i][j] == '1')
{
ret++;
bfs(grid, i, j); //标记联通1
}
}
}
return ret;
}
void bfs(vector<vector<char>>& grid, int i, int j)
{
queue<pair<int, int>> q;
q.push({i, j});
vis[i][j] = true;
while(q.size())
{
auto [a, b] = q.front();
q.pop();
for(int k = 0; k < 4; k++)
{
int x = a + dx[k], y = b + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == '1')
{
q.push({x, y});
vis[x][y] = true;
}
}
}
}
};3.岛屿的最大面积
岛屿的最大面积
dfs解决代码：
class Solution {
public:
bool vis[51][51];
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
int count;
int maxAreaOfIsland(vector<vector<int>>& grid) {
m = grid.size(), n = grid[0].size();
int ret = 0;
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if(!vis[i][j] && grid[i][j] == 1)
{
count = 0;
dfs(grid, i, j);
ret = max(ret, count);
}
}
}
return ret;
}
void dfs(vector<vector<int>>& grid, int i, int j)
{
count++;
vis[i][j] = true;
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1)
{
dfs(grid, x, y);
}
}
}
};bfs解决代码：
class Solution {
public:
bool vis[51][51];
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
int maxAreaOfIsland(vector<vector<int>>& grid) {
m = grid.size(), n = grid[0].size();
int ret = 0;
for(int i = 0; i < m; i++)
{
for(int j = 0; j < n; j++)
{
if(!vis[i][j] && grid[i][j] == 1)
{
ret = max(ret, bfs(grid, i, j));
}
}
}
return ret;
}
int bfs(vector<vector<int>>& grid, int i, int j)
{
int count = 0;
queue<pair<int, int>> q;
q.push({i, j});
vis[i][j] = true;
count++;
while(!q.empty())
{
auto [a, b] = q.front();
q.pop();
for(int k = 0; k < 4; k++)
{
int x = a + dx[k], y = b + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1)
{
q.push({x, y});
vis[x][y] = true;
count++;
}
}
}
return count;
}
};4.被围绕的区域
被围绕的区域
dfs解决代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
void solve(vector<vector<char>>& board) {
m = board.size(), n = board[0].size();
for(int j = 0; j < n; j++)
{
if(board[0][j] == 'O') dfs(board, 0, j);
if(board[m - 1][j] == 'O') dfs(board, m - 1, j);
}
for(int i = 0; i < m; i++)
{
if(board[i][0] == 'O') dfs(board, i, 0);
if(board[i][n - 1] == 'O') dfs(board, i, n - 1);
}
for(int i = 0; i < m; i++) //还原
for(int j = 0; j < n; j++)
if(board[i][j] == 'O')
board[i][j] = 'X';
else if(board[i][j] == '.')
board[i][j] = 'O';
}
void dfs(vector<vector<char>>& board, int i, int j)
{
board[i][j] = '.';
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
{
dfs(board, x, y);
}
}
}
};bfs解决代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
void solve(vector<vector<char>>& board) {
m = board.size(), n = board[0].size();
for(int j = 0; j < n; j++)
{
if(board[0][j] == 'O') bfs(board, 0, j);
if(board[m - 1][j] == 'O') bfs(board, m - 1, j);
}
for(int i = 0; i < m; i++)
{
if(board[i][0] == 'O') bfs(board, i, 0);
if(board[i][n - 1] == 'O') bfs(board, i, n - 1);
}
for(int i = 0; i < m; i++) //还原
for(int j = 0; j < n; j++)
if(board[i][j] == 'O')
board[i][j] = 'X';
else if(board[i][j] == '.')
board[i][j] = 'O';
}
void bfs(vector<vector<char>>& board, int i, int j)
{
queue<pair<int, int>> q;
q.push({i, j});
board[i][j] = '.';
while(!q.empty())
{
auto [a, b] = q.front();
q.pop();
for(int k = 0; k < 4; k++)
{
int x = a + dx[k], y = b + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
{
q.push({x, y});
board[x][y] = '.';
}
}
}
}
};5.太平洋大西洋水流问题
太平洋大西洋水流问题
代码：
class Solution {
public:
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int m, n;
vector<vector<int>> pacificAtlantic(vector<vector<int>>& h) {
m = h.size(), n = h[0].size();
vector<vector<bool>> pac(m, vector<bool>(n));
vector<vector<bool>> atl(m, vector<bool>(n));
//pac -- 太平洋
for(int i = 0; i < m; i++) dfs(h, i, 0, pac);
for(int j = 0; j < n; j++) dfs(h, 0, j, pac);
//atl -- 大西洋
for(int i = 0; i < m; i++) dfs(h, i, n - 1, atl);
for(int j = 0; j < n; j++) dfs(h, m - 1, j, atl);
vector<vector<int>> ret;
for(int i = 0; i < m; i++)
for(int j = 0; j < n; j++)
if(pac[i][j] && atl[i][j])
ret.push_back({i, j});
return ret;
}
void dfs(vector<vector<int>>& h, int i, int j, vector<vector<bool>>& vis)
{
vis[i][j] = true;
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && h[x][y] >= h[i][j])
{
dfs(h, x, y, vis);
}
}
}
};6.扫雷游戏
扫雷游戏
代码：
class Solution {
public:
int dx[8] = {0, 0, -1, -1, -1, 1, 1, 1};
int dy[8] = {-1, 1, -1, 0, 1, -1, 0, 1};
int m, n;
vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
m = board.size(), n = board[0].size();
int x = click[0], y = click[1];
if(board[x][y] == 'M')
{
board[x][y] = 'X';
return board;
}
dfs(board, x, y);
return board;
}
void dfs(vector<vector<char>>& board, int i, int j)
{
//统计一下周围地雷个数
int count = 0;
for(int k = 0; k < 8; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M')
count++;
}
if(count)//如果周围有地雷记录地雷数
{
board[i][j] = count + '0';
return;
}
else//周围无地雷继续向外扩展
{
board[i][j] = 'B';
for(int k = 0; k < 8; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E')
{
dfs(board, x, y);
}
}
}
}
};7.衣橱整理
衣橱整理
代码：
class Solution {
public:
int m, n, cnt;
int ret;
bool vis[101][101];
int dx[4] = {0, 0, -1, 1};
int dy[4] = {-1, 1, 0, 0};
int wardrobeFinishing(int _m, int _n, int _cnt) {
m = _m, n = _n, cnt = _cnt;
dfs(0, 0);
return ret;
}
void dfs(int i, int j)
{
ret++;
vis[i][j] = true;
for(int k = 0; k < 4; k++)
{
int x = i + dx[k], y = j + dy[k];
if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && check(x, y))
{
dfs(x, y);
}
}
}
bool check(int i, int j)
{
int sum = 0;
while(i)
{
sum += i % 10;
i /= 10;
}
while(j)
{
sum += j % 10;
j /= 10;
}
return sum <= cnt;
}
};
