@(labuladong的算法小抄)[DFS, 并查集]
leetcode 130. 被圍繞的區域
題目描述
解題思路
參考:labuladong的算法小抄P407
DFS
時間複雜度o(mn)
class Solution {
public void solve(char[][] board) {
if (board.length == 0) return;
int m = board.length, n = board[0].length;
/* 将與第一行和最後一行關聯的'0'全變成'#' */
for (int j = 0; j < n; j++) {
dfs(board, 0, j);
dfs(board, m - 1, j);
}
/* 将與第一列和最後一列關聯的'0'全變成'#' */
for (int i = 0; i < m; i++) {
dfs(board, i, 0);
dfs(board, i, n - 1);
}
/* 周遊整個棋盤 */
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
/* 剩下的所有'0'都是應該替換成'X'的 */
if (board[i][j] == 'O') {
board[i][j] = 'X';
}
/* 把所有'#'恢複為‘O’ */
if (board[i][j] == '#') {
board[i][j] = 'O';
}
}
}
}
/* 從board[row][col]處開始DFS,将所有相鄰的'O'變為'#' */
private void dfs(char[][] board, int row, int col) {
int m = board.length, n = board[0].length;
/* 越界直接傳回 */
if (row < 0 || row >= m || col < 0 || col >= n) {
return;
}
if (board[row][col] != 'O') {
return;
}
/* 進行替換 */
board[row][col] = '#';
/* 向四周DFS搜尋 */
dfs(board, row + 1, col);
dfs(board, row - 1, col);
dfs(board, row, col + 1);
dfs(board, row, col - 1);
}
}
并查集
class Solution {
public void solve(char[][] board) {
if (board.length == 0) return;
int m = board.length, n = board[0].length;
/* 給虛拟節點dummy留一個位置 */
UnionFind uf = new UnionFind(m * n + 1);
/* 隻要和dummy連通的'O'都不需要替換 */
int dummy = m * n;
/* 将第一行和最後一行的'O'與dummy連通 */
for (int j = 0; j < n; j++) {
if (board[0][j] == 'O')
uf.union(j, dummy);
if (board[m - 1][j] == 'O')
uf.union((m - 1) * n + j, dummy);
}
/* 将第一列和最後一列的'O'與dummy連通 */
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O')
uf.union(i * n, dummy);
if (board[i][n - 1] == 'O')
uf.union(i * n + n - 1, dummy);
}
/* 方向數組是搜尋上下左右四個方向的常用手法 */
int[][] d = new int[][]{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
for (int i = 1; i < m - 1; i++) {
for (int j = 1; j < n - 1; j++) {
if (board[i][j] == 'O') {
/* 将此'O'和上下所有的'O'連通 */
for (int k = 0; k < 4; k++) {
int row = i + d[k][0];
int col = j + d[k][1];
if (board[row][col] == 'O')
uf.union(n * i + j, row * n + col);
}
}
}
}
/* 現在,沒有被X包圍的O都和dummy連通了,剩下的不和dummy連通的'O'都要替換 */
/* 注意,一定要等上面的for循環結束後再替換,不能邊判斷邊替換! */
for (int i = 1; i < m - 1; i++) {
for (int j = 1; j < n - 1; j++) {
if (!uf.isConnected(n * i + j, dummy))
board[i][j] = 'X';
}
}
}
/* 并查集 */
class UnionFind {
/* 連通分量個數 */
private int count;
/* 存儲每個節點的父節點 */
private int[] parent;
/* 記錄每棵樹的重量 */
private int[] size;
public UnionFind(int n) {
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
/* 将兩個節點連通 */
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;
/* 小樹接到大樹下面,保證平衡 */
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootP] += size[rootQ];
} else {
parent[rootQ] = rootP;
size[rootQ] += size[rootP];
}
count--;
}
/* 判斷兩個節點是否連通 */
public boolean isConnected(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
return rootP == rootQ;
}
/* 尋找節點x的根節點 */
public int find(int x) {
/* 根節點的父節點是它自己 */
while (parent[x] != x) {
/* 進行路徑壓縮 */
parent[x] = parent[parent[x]];
x = parent[x];
}
return x;
}
/* 傳回并查集中連通分量的個數 */
public int count() {
return count;
}
}
}