leetcode算法题之floodfill算法---深搜(dfs)-编程知识

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
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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
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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
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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
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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
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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
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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
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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;}
};