文章目录
- 太平洋大西洋流水问题
- 扫雷游戏
- 迷路的机器人
太平洋大西洋流水问题
class Solution {
public:
vector<vector<int>> res;
int m = 0, n = 0;
vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
m = heights.size();
n = heights[0].size();
vector<vector<bool>> check1(m, vector<bool>(n, false));
vector<vector<bool>> check2(m, vector<bool>(n, false));
for (int i = 0; i < m; ++i) {
if (!check1[i][0])
dfs(heights, check1, i, 0);
if (!check2[i][n - 1])
dfs(heights, check2, i, n - 1);
}
for (int i = 0; i < n; ++i) {
if (!check1[0][i])
dfs(heights, check1, 0, i);
if (!check2[m - 1][i])
dfs(heights, check2, m - 1, i);
}
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
if (check1[i][j] && check2[i][j])
res.push_back({i, j});
}
return res;
}
void dfs(vector<vector<int>>& heights, vector<vector<bool>>& check, int i, int j) {
check[i][j] = true;
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
for(int k = 0; k < 4; k++){
int x = i + dx[k];
int y = j + dy[k];
if(x >= 0 && y >= 0 && x < m && y < n && heights[x][y] >= heights[i][j] && !check[x][y])
dfs(heights, check, x, y);
}
}
};
扫雷游戏
代码1 与 代码2 的相似度达到 99%，但其差异性在逻辑思维上存在先后影响。有兴趣的道友可以细细体会一下
代码1：
class Solution {
public:
int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
bool check[51][51] = {false};
int m = 0, n = 0;
vector<vector<char>> updateBoard(vector<vector<char>>& board,
vector<int>& click) {
// 预处理,将已经处理过的点标记
m = board.size();
n = board[0].size();
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
if (board[i][j] != 'M' && board[i][j] != 'E')
check[i][j] = true;
}
// 解题默认点击未挖出的点：未挖出的点分为两种情况：M & E
if (board[click[0]][click[1]] == 'M') {
board[click[0]][click[1]] = 'X';
} else if (board[click[0]][click[1]] == 'E') {
dfs(board, click[0], click[1]);
}
return board;
}
void dfs(vector<vector<char>>& board, int i, int j) {
board[i][j] = 'B';
int count = 0;
for (int k = 0; k < 8; ++k) {
int x = i + dx[k];
int y = j + dy[k];
if (x >= 0 && y >= 0 && x < m && y < n && board[x][y] == 'M')
count++;
}
if (count != 0)
board[i][j] = '0' + count;
else {
for (int k = 0; k < 8; ++k) {
int x = i + dx[k];
int y = j + dy[k];
if (x >= 0 && y >= 0 && x < m && y < n && !check[x][y] && board[x][y] == 'E')
dfs(board, x, y);
}
}
// 处理过的点进行标记
check[i][j] = true;
}
};
代码2：
class Solution {
public:
int dx[8] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[8] = {-1, 0, 1, -1, 1, -1, 0, 1};
bool check[51][51] = {false};
int m = 0, n = 0;
vector<vector<char>> updateBoard(vector<vector<char>>& board,
vector<int>& click) {
// 预处理,将已经处理过的点标记
m = board.size();
n = board[0].size();
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
if (board[i][j] != 'M' && board[i][j] != 'E')
check[i][j] = true;
}
// 解题默认点击未挖出的点：未挖出的点分为两种情况：M & E
if (board[click[0]][click[1]] == 'M') {
board[click[0]][click[1]] = 'X';
} else if (board[click[0]][click[1]] == 'E') {
dfs(board, click[0], click[1]);
}
return board;
}
void dfs(vector<vector<char>>& board, int i, int j) {
board[i][j] = 'B';
int count = 0;
for (int k = 0; k < 8; ++k) {
int x = i + dx[k];
int y = j + dy[k];
if (x >= 0 && y >= 0 && x < m && y < n && board[x][y] == 'M')
count++;
}
// 处理过的点进行标记
check[i][j] = true;
if (count != 0)
board[i][j] = '0' + count;
else {
for (int k = 0; k < 8; ++k) {
int x = i + dx[k];
int y = j + dy[k];
if (x >= 0 && y >= 0 && x < m && y < n && !check[x][y])
dfs(board, x, y);
}
}
}
};
迷路的机器人
class Solution {
public:
vector<vector<int>> res;
int m = 0, n = 0;
bool check[101][101] = {false};
vector<vector<int>> pathWithObstacles(vector<vector<int>>& obst) {
m = obst.size();
n = obst[0].size();
if (obst[0][0] == 1 || obst[m - 1][n - 1] == 1)
return {};
res.push_back({0, 0});
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
if (obst[i][j] == 1)
check[i][j] = true;
}
if (!dfs(obst, 0, 0))
res.pop_back();
return res;
}
bool dfs(vector<vector<int>>& obst, int i, int j) {
if (res.size() == m + n - 1) {
return true;
}
int dx[2] = {0, 1};
int dy[2] = {1, 0};
for (int k = 0; k < 2; k++) {
int x = i + dx[k];
int y = j + dy[k];
if (x < m && y < n && !check[x][y]) {
if (obst[x][y] == 0) {
res.push_back({x, y});
if (dfs(obst, x, y))
return true;
else {
res.pop_back();
check[x][y] = true;
}
}
}
}
return false;
}
};
