算法作业HW19 130. Surrounded Regions

Description:

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

 

Note:

For example,

X X X XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

 

Solution:

  Analysis and Thinking:

题目意思为给定一个二维矩阵，里面包含O、X，要求把所有被X包围的O变为X，题目可以用很多解法比如DFS、BFS等，这里采用dfs的方式进行解决

 

  Steps:

  1.判断输入的board矩阵是否为空，若是则返回

2.获取矩阵的行列长度

3.按行元素进行深度优先遍历，判断输入矩阵的下标信息，当行列坐标都没越界，元素置为‘D’

4.按列元素进行深度优先遍历，判断输入矩阵的下标信息，当行列坐标都没越界，元素置为‘D’

5.深度优先的顺序为某一元素的四周，即其行列坐标加减1可得的四个位置的元素

6.最后循环遍历后的board，当其元素为‘O’，置换为‘X’，返回

Codes:

public class Solution {    int i,j;    char[][] board;        void dfs(int x, int y){        if(x<0 || x>=i || y<0 || y>=j || board[x][y]!='O') return;        board[x][y]='D';        dfs(x-1,y);        dfs(x+1,y);        dfs(x,y-1);        dfs(x,y+1);    }        public void methodMain(char[][] board){        if(board==null || board.length==0) return;        this.board=board;        i=board.length;        j=board[0].length;                for(int y=0;y<i;y++){            dfs(0,y);            dfs(y-1,j);        }                for(int x=1;x<j-1;x++){            dfs(x,0);            dfs(x,i-1);        }                for(int x=0;x<m;x++)            for(int y=0;y<n;y++){                if(board[x][y]=='O') board[x][y]='X';                else if(board[x][y]=='D') board[i][j]='O';            }      }}

Results: