leetcode算法题之floodfill算法---深搜(dfs)
本章目录
- 1.图像渲染
- 2.岛屿数量
- 3.岛屿的最大面积
- 4.被围绕的区域
- 5.太平洋大西洋水流问题
- 6.扫雷游戏
- 7.机器人的运动范围
1.图像渲染
图像渲染
class Solution {
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int newColor,prev;
int m,n;
public:
vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
newColor = color;
prev = image[sr][sc];
if(color == image[sr][sc]) return image;
m = image.size(),n = image[0].size();
dfs(image,sr,sc);
return image;
}
void dfs(vector<vector<int>>& image,int i,int j)
{
image[i][j] = newColor;
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);
}
}
}
};
2.岛屿数量
岛屿数量
class Solution {
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
bool vis[301][301];
int m,n;
public:
int numIslands(vector<vector<char>>& 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++;
dfs(grid,i,j);//把这块区域都标记成已访问
}
}
}
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);
}
}
}
};
3.岛屿的最大面积
岛屿的最大面积
class Solution {
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int m,n,count;
bool vis[51][51];
public:
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);
}
}
}
};
4.被围绕的区域
被围绕的区域
class Solution {
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int m,n;
public:
//正难则反：先处理与边缘'O'相连的连通块
void solve(vector<vector<char>>& board) {
m = board.size(),n = board[0].size();
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 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);
}
//还原
for(int i=0;i<m;i++)
{
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);
}
}
}
};
5.太平洋大西洋水流问题
太平洋大西洋水流问题
class Solution {
//正难则反：求出哪些点能流进太平洋，哪些点能流进大西洋，在求之间的交集即可
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int m,n;
vector<vector<bool>> pac;
vector<vector<bool>> alt;
vector<vector<int>> ret;
public:
vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
m = heights.size(),n = heights[0].size();
pac = vector<vector<bool>>(m,vector<bool>(n));
alt = vector<vector<bool>>(m,vector<bool>(n));
for(int i=0;i<m;i++)
{
dfs(heights,i,0,pac);
dfs(heights,i,n-1,alt);
}
for(int j=0;j<n;j++)
{
dfs(heights,0,j,pac);
dfs(heights,m-1,j,alt);
}
//判断
for(int i=0;i<m;i++)
{
for(int j=0;j<n;j++)
{
if(pac[i][j] && alt[i][j])
{
ret.push_back({i,j});
}
}
}
return ret;
}
void dfs(vector<vector<int>>& heights,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] && heights[x][y]>=heights[i][j])
{
dfs(heights,x,y,vis);
}
}
}
};
6.扫雷游戏
扫雷游戏
class Solution {
int dx[8] = {0,0,1,-1,1,1,-1,-1};
int dy[8] = {1,-1,0,0,1,-1,1,-1};
int m,n;
public:
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)
{
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)
{
board[i][j] = count +'0';
return;
}
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.机器人的运动范围
机器人的运动范围
class Solution {
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
int m,n,k;
int ret =0;
bool vis[101][101];
public:
//从(0，0)点，进行一次深度优先遍历即可
int movingCount(int _k, int _m, int _n) {
k = _k,m=_m,n=_n;
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 x,int y)
{
int tmp = 0;
while(x)
{
tmp += x%10;
x /= 10;
}
while(y)
{
tmp += y%10;
y /= 10;
}
return tmp<=k;
}
};
这个系列就到处结束啦，希望对大家有所帮助！