UnionFInd-----130. Surrounded Regions

原题目

着色法，把陆地圈出来然后再做遍历。但是出现了栈溢出的情况。

1.栈溢出

import java.util.ArrayList;import java.util.List;/** * Created by XQF on 2017/8/11. */class ListNode {    int val;    ListNode next;    public ListNode(int a) {        this.val = a;    }}class Point {    int x;    int y;    public Point(int x, int y) {        this.x = x;        this.y = y;    }    @Override    public String toString() {        return x + " " + y + "\n";    }}public class Solution {    private int[][] next = {            {0, 1},            {1, 0},            {0, -1},            {-1, 0}    };    public void solve(char[][] board) {        if (board == null || board.length == 0) {            return;        }        int len1 = board.length;        int len2 = board[0].length;        int[][] book = new int[len1][len2];        for (int i = 0; i < len1; i++) {            for (int j = 0; j < len2; j++) {                book[i][j] = 0;            }        }        List<List<Point>> points = new ArrayList<>();        for (int i = 0; i < len1; i++) {            for (int j = 0; j < len2; j++) {                if (board[i][j] == 'O' && book[i][j] != 1) {                    List<Point> temp = new ArrayList<>();                    book[i][j] = 1;                    Point p = new Point(i, j);                    temp.add(p);                    dfs(board, book, i, j, temp);                    points.add(temp);                }            }        }        for (List<Point> list : points) {            if (isSurranded(list, len1, len2)) {                help(list, board);            }        }    }    private void dfs(char[][] board, int[][] book, int i, int j, List<Point> temp) {        int len1 = board.length;        int len2 = board[0].length;        for (int k = 0; k <= 3; k++) {            int x = i + next[k][0];            int y = j + next[k][1];            if (x < 0 || x >= len1 || y < 0 || y >= len2) {                continue;            }            if (board[x][y] == 'O' && book[x][y] == 0) {                book[x][y] = 1;                Point p = new Point(x, y);                temp.add(p);                dfs(board, book, x, y, temp);            }        }    }    public boolean isSurranded(List<Point> list, int x, int y) {        boolean result = true;        for (int i = 0; i < list.size(); i++) {            Point p = list.get(i);            if (p.x + 1 >= x || p.y + 1 >= y) {                result = false;                break;            }            if (p.x - 1 < 0 || p.y - 1 < 0) {                result = false;                break;            }        }        return result;    }    public void help(List<Point> list, char[][] board) {        for (Point p : list) {            int x = p.x;            int y = p.y;            board[x][y] = 'X';        }    }    public void print(char[][] board) {        for (int i = 0; i < board.length; i++) {            for (int j = 0; j < board[i].length; j++) {                System.out.print(board[i][j] + " ");            }            System.out.println();        }    }    public static void main(String[] args) {        Solution solution = new Solution();        char[][] board = new char[][]{                {'X', 'O', 'X', 'X'},                {'X', 'O', 'O', 'X'},                {'X', 'X', 'O', 'X'},                {'X', 'O', 'X', 'X'},        };        solution.solve(board);        solution.print(board);    }}

2.队列

参考了别人的思路

既然说我栈溢出，那我就不用栈我用队列哈哈。使用宽度优先。但是为了减少搜索的次数，根据题目意思进行简单的改变而不是全部搜索（将道理要是深搜也这样优化应该也是能通过的）。我们只需要根据四个边的情况来搜索，一旦出现‘O‘进行扩展并做好标记。这样就好做多了

 public int[][] next = new int[][]{            {1, 0},            {0, -1},            {-1, 0},            {0, 1}    };    public void solve(char[][] board) {        if (board == null || board.length == 0) {            return;        }        int len1 = board.length;        int len2 = board[0].length;        int[][] book = new int[len1][len2];        for (int i = 0; i < len1; i++) {            bfs(board, book, i, 0);            bfs(board, book, i, len2 - 1);        }        for (int i = 0; i < len2; i++) {            bfs(board, book, 0, i);            bfs(board, book, len1 - 1, i);        }        for (int i = 0; i < len1; i++) {            for (int j = 0; j < len2; j++) {                if (board[i][j] == 'O') {                    board[i][j] = 'X';                } else if (board[i][j] == '#') {                    board[i][j] = 'O';                }            }        }    }    private void bfs(char[][] board, int[][] book, int i, int j) {        if (board[i][j] != 'O') {            return;        }        int len1 = board.length;        int len2 = board[0].length;        Queue q = new LinkedList<>();        board[i][j] = '#';        q.add(new Point(i, j));        while (!q.isEmpty()) {            Point p = (Point) q.poll();            int x = p.x;            int y = p.y;            for (int k = 0; k <= 3; k++) {                int xx = x + next[k][0];                int yy = y + next[k][1];                if (xx < 0 || xx >= len1 || yy < 0 || yy >= len2) {                    continue;                }                if (board[xx][yy] == 'O' && book[xx][yy] == 0) {                    book[xx][yy] = 1;                    board[xx][yy] = '#';                    q.add(new Point(xx, yy));                }            }        }    }