【LeetCode】 Surrounded Regions （BFS && DFS）

標籤：dfs   bfs   leetcode   

題目：Surrounded Regions

廣搜和深搜都能解決，但是LeetCode上使用深搜時會棧溢出

DFS:

<span style="font-size:18px;">/*LeetCode Surrounded Regions * 題目：給定一個字元數組，由'X'和'O'組成，找到所有被x包圍的o並將其替換為x * 思路：只要替換被包圍的o就行，如果有一個o是邊界或者上下左右中有一個是o且這個o不會被替換，則該點也不會被替換 * 從四條邊開始，因為在這4周的一定不是被包圍的所以用他們開始找到廣搜的隊列，如果隊列為空白，那麼就是所有的o都被包圍 */package javaTrain;public class Train25 {public static void solve(char[][] board) {long n = board.length;if(n==0) return;long m = board[0].length; for(long i = 0;i < m;i++){//對第一行和最後一行的字元進行廣搜bfs(board,0,i);bfs(board,n-1,i);}for(long j = 1;j < n-1;j++){ //對第一列和最後一列的字元進行廣搜，去除4條邊重複的字元bfs(board,j,0);bfs(board,j,m-1);}for(int i = 0;i < n;i++){for(int j = 0;j < m;j++){if(board[i][j] == 'O') board[i][j] = 'X';//被包圍的o需取代else if(board[i][j] == '$') board[i][j] = 'O';//標記的不被包圍的o保持原樣}}}private static void bfs(char[][] board,int i,int j){long n = board.length;long m = board[0].length;if(i < 0 || i>=n||j<0||j>=m||board[i][j] != 'O') return; //邊界的點都不被包圍 board[i][j] = '$'; bfs(board,i-1,j);bfs(board,i,j-1);bfs(board,i+1,j);bfs(board,i,j+1); } public static void main(String args[]){char board[][] = {{'O','X','O'},{'X','O','X'},{'O','X','O'}};solve(board); for(int i = 0;i < board.length;i++){for(int j = 0;j < board[0].length;j++){ System.out.print(board[i][j]);}System.out.println();} }}</span>

BFS:

<span style="font-size:18px;">// LeetCode, Surrounded Regions// BFS，時間複雜度O(n)，空間複雜度O(n)class Solution {public:    void solve(vector<vector<char>> &board) {        if (board.empty()) return;        const int m = board.size();        const int n = board[0].size();        for (int i = 0; i < n; i++) {            bfs(board, 0, i);            bfs(board, m - 1, i);        }        for (int j = 1; j < m - 1; j++) {            bfs(board, j, 0);            bfs(board, j, n - 1);        }        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';    }private:    void bfs(vector<vector<char>> &board, int i, int j) {        typedef pair<int, int> state_t;        queue<state_t> q;        const int m = board.size();        const int n = board[0].size();        auto is_valid = [&](const state_t &s) {            const int x = s.first;            const int y = s.second;            if (x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O')                return false;            return true;        };        auto state_extend = [&](const state_t &s) {            vector<state_t> result;            const int x = s.first;            const int y = s.second;            // 上下左右            const state_t new_states[4] = {{x-1,y}, {x+1,y},                    {x,y-1}, {x,y+1}};            for (int k = 0; k < 4;  ++k) {                if (is_valid(new_states[k])) {                    // 既有標記功能又有去重功能                    board[new_states[k].first][new_states[k].second] = '+';                    result.push_back(new_states[k]);                }            }            return result;        };        state_t start = { i, j };        if (is_valid(start)) {            board[i][j] = '+';            q.push(start);        }        while (!q.empty()) {            auto cur = q.front();            q.pop();            auto new_states = state_extend(cur);            for (auto s : new_states) q.push(s);        }    }};</span>

【LeetCode】 Surrounded Regions （BFS && DFS）