"Leetcode" surrounded regions (BFS && DFS)

Title: Surrounded regions

Wide search and deep search can be solved, but leetcode on the use of deep search will overflow the stack

Dfs:

<span style= "FONT-SIZE:18PX;" >/*leetcode surrounded Regions * Title: Given an array of characters, consisting of ' X ' and ' O ', find all o surrounded by x and replace it with X-ideas: Just replace the surrounded O-line, If there is an O that is the boundary or if one of the upper and lower left is O and this o is not replaced, then the point will not be replaced * starting from four sides, because in this 4 weeks must not be surrounded so use them to start to find the wide search queue, if the queue is empty, then all O is surrounded */package Javatrain;public class Train25 {public static void solve (char[][] board) {Long n = board.length;if (n==0) Return;long m = b Oard[0].length; for (Long i = 0;i < m;i++) {//The first and last line characters are extensively searched BFS (board,0,i); BFS (board,n-1,i);} for (Long j = 1;j < n-1;j++) {//wide search for characters in the first and last columns, removing 4-edged duplicate characters BFS (board,j,0); BFS (board,j,m-1); for (int i = 0;i < n;i++) {if (int j = 0;j < m;j++) {if (board[i][j] = = ' O ') board[i][j] = ' X ';//The enclosing O is required to replace else if (board[i ][J] = = ' $ ') board[i][j] = ' o ';//marked non-enclosed O remains as-is}}}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; The points of the boundary are not surrounded by 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 &L T 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, time complexity O (n), Space complexity O (N) class Solution {Public:void solve (vector<vector<ch        Ar>> &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> STA        te_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;            Up or down 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])) {//both tagged and re-function board[new_state                    S[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)