深度優先搜索在得到一個新節點時立即對新節點進行遍歷
從一個節點出發，使用 DFS 對一個圖進行遍歷時，能夠遍歷到的節點都是從初始節點可達的，DFS 常用來求解這種 可達性 問題。
在程序實現 DFS 時需要考慮以下問題：
棧：用棧來保存當前節點信息，當遍歷新節點返回時能夠繼續遍歷當前節點。可以使用遞歸棧。
標記：和 BFS 一樣同樣需要對已經遍歷過的節點進行標記。
1. 查找最大的連通面積
/*
* 題目：最大連通的面積
* */
public int maxAreaOfIsland(int[][] grid) {
int m = grid.length, n = grid[0].length;
int book[][] = new int[m][n];
int maxArea = 0;
int direction[][] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
maxArea = Math.max(maxArea, connectArea(grid, i, j, book, direction));
}
}
return maxArea;
}
private int connectArea(int[][] grid, int i, int j, int[][] book, int[][] direction) {
if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) return 0;
if (book[i][j] == 1 || grid[i][j] == 0) return 0;
book[i][j] = 1;
int count = 1;
for (int k = 0; k < direction.length; k++) {
int x = i + direction[k][0], y = j + direction[k][1];
count += connectArea(grid, x, y, book, direction);
}
return count;
}
2. 矩陣中的連通分量數目
/*
*矩陣中的連通分量數目
* */
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length, n = grid[0].length;
int count = 0;
int direction[][] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == '1') {
count += 1;
connectArea1(grid, i, j, direction);
}
}
}
return count;
}
private void connectArea1(char[][] grid, int i, int j, int[][] direction) {
if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) return;
if (grid[i][j] == '0') return;
grid[i][j] = '0';
for (int k = 0; k < direction.length; k++) {
int x = i + direction[k][0], y = j + direction[k][1];
connectArea1(grid, x, y, direction);
}
return;
}
3. 好友關係的連通分量數目
/*
* 好友關係的連通分量數目
* 題目描述：好友關係可以看成是一個無向圖，例如第 0 個人與第 1 個人是好友，那麼 M[0][1] 和 M[1][0] 的值都爲 1。
* 解析：注意匹配規則！
* */
public int findCircleNum(int[][] M) {
if (M == null || M.length == 0) {
return 0;
}
int m = M.length, n = M[0].length;
int count = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j <= i; j++) {
if (M[i][j] == 1) {
count += 1;
connectArea2(M, i, j);
}
}
}
return count;
}
private void connectArea2(int[][] M, int i, int j) {
if (i < 0 || j < 0 || i >= M.length || j > M[0].length) return;
if (M[i][j] == 0 && M[j][i] == 0) return;
M[i][j] = 0;
M[j][i] = 0;
for (int k = 0; k < M.length; k++) {
connectArea2(M, i, k);
connectArea2(M, k, j);
}
return;
}
4. 填充封閉區域
/*
* 填充封閉區域
* 題目描述：使被 'X' 包圍的 'O' 轉換爲 'X'。
* */
public void solve(char[][] board) {
if (board == null || board.length == 0) {
return;
}
int m = board.length, n = board[0].length;
int direction[][] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
for (int i = 0; i < m; i++) {
connectArea3(board, i, 0, direction);
connectArea3(board, i, n - 1, direction);
}
for (int j = 0; j < n; j++) {
connectArea3(board, 0, j, direction);
connectArea3(board, m - 1, j, direction);
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] == 'O') {
board[i][j] = 'X';
}
if (board[i][j] == 'Z') {
board[i][j] = 'O';
}
}
}
}
private void connectArea3(char[][] grid, int i, int j, int[][] direction) {
if (i < 0 || j < 0 || i >= grid.length || j >= grid[0].length) return;
if (grid[i][j] != 'O') return;
grid[i][j] = 'Z';
for (int k = 0; k < direction.length; k++) {
int x = i + direction[k][0], y = j + direction[k][1];
connectArea3(grid, x, y, direction);
}
return;
}
5. 能到達的太平洋和大西洋的區域
/*
* 能到達的太平洋和大西洋的區域
* 左邊和上邊是太平洋，右邊和下邊是大西洋，內部的數字代表海拔，海拔高的地方的水能夠流到低的地方，求解水能夠流到太平洋和大西洋的所有位置。
* */
public List<List<Integer>> pacificAtlantic(int[][] matrix) {
List<List<Integer>> list = new LinkedList<>();
if (matrix == null || matrix.length == 0) return list;
int m = matrix.length, n = matrix[0].length;
int direction[][] = {{0, 1}, {0, -1}, {-1, 0}, {1, 0}};
int bookT[][] = new int[m][n];//表示太平洋可達性記錄
int bookD[][] = new int[m][n];//表示大西洋可達性記錄
for (int i = 0; i < m; i++) {
arrive(matrix, i, 0, direction, bookT);
arrive(matrix, i, n - 1, direction, bookD);
}
for (int j = 0; j < n; j++) {
arrive(matrix, 0, j, direction, bookT);
arrive(matrix, m - 1, j, direction, bookD);
}
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (bookD[i][j] == 1 && bookT[i][j] == 1) {
List<Integer> curElem = new LinkedList<>(Arrays.asList(i, j));
list.add(curElem);
}
}
}
return list;
}
public void arrive(int[][] matrix, int i, int j, int direction[][], int book[][]) {
if (book[i][j] != 0) return;
book[i][j] = 1;//代表該點可達
for (int k = 0; k < direction.length; k++) {
int x = i + direction[k][0];
int y = j + direction[k][1];
if (x < 0 || x >= matrix.length || y < 0 || y >= matrix[0].length) continue;
if (matrix[x][y] >= matrix[i][j]) {
arrive(matrix, x, y, direction, book);
}
}
}