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Topic description
You are given an mxn matrix board, consisting of several characters 'X' and 'O', find all areas surrounded by 'X', and fill all 'O's in these areas with 'X'.
Example 1:
Input: board = [["X","X","X","X"],["X","O","O","X"],["X","X", "O","X"],["X","O","X","X"]] Output: [["X","X","X","X"],[" X","X","X","X"],["X","X","X","X"],["X","O","X","X" ]] Explanation: The enclosed interval will not exist on the boundary, in other words, the 'O' on any boundary will not be filled with 'X'. Any 'O' that is not on a boundary, or is not connected to an 'O' on a boundary, will eventually be filled with an 'X'. Two elements are said to be "connected" if they are adjacent horizontally or vertically. Example 2:
Input: board = [["X"]] Output: [["X"]]
hint:
- m == board.length
- n == board[i].length
- 1 <= m, n <= 200
- board[i][j] is 'X' or 'O'
Problem solving ideas
给定一个二维数组board，数组中元素只包含'O','X'。将二位数数组中所有不和边界'O'直接或间接关联的'O'元素替换成'X'。
根据题意，这道题的重点就是要找出所有与边界'0'直接或间接关联的点；
比如点(1,0)为'0'，则与它直接关联的点有(2,0),(1,1),(0,0)。即对原坐标进行四个方向的加减(1,0),(-1,0),(0,1),(0,-1);
即上下左右四个方向，然后这几个点又有直接关联的上下左右四个点，依次类推，直到这些关联的点元素不是'O';
所以解这道题有深度优先和广度优先两种选择；
深度优先
遍历上、下、左、右边界的元素，每一次找一个方向上所有的直接或间接与'O'关联的元素，将它们替换成其它元素，比如'B';
遍历完成之后将原数组进行遍历，找到不是'B'且是'O'的元素，将它们设置成'X',然后再将元素值为'B'的替换成'O';
广度优先
第一层遍历获取上、下、左、右边界上所有的'O'元素，将它们设置成其它的元素，比如'B'，将它放入队列queue中；
遍历queue，截止条件为queue为空，每一次循环从queue中弹出栈顶元素，然后对原坐标进行四个方向的加减(1,0),(-1,0),(0,1),(0,-1)；
获取与当前元素直接关联的节点，判断如果计算出来的位置的元素为'O'，则将这个节点的位置放入queue中，并且将该元素修改为'B';
When the content of the queue is empty, that is, all elements in the two-dimensional array that are directly or indirectly related to the boundary 'O' are found. At this time, the two-dimensional array is traversed, and the element value is not 'B' and the element value is 'O'. Replace the element with 'X', and then replace the element value with 'B' with 'O';
Code
public void solve(char[][] board) {
n = board.length;
m = board[0].length;
// 深度优先
// 第一层左右边界
for (int i=0; i < n;i ++) {
// 第一列
dfs(board,i,0);
// 末尾列
dfs(board,i,m-1);
}
// 上下边界
for (int i=0; i < m;i ++) {
// 第一列
dfs(board,0,i);
// 末尾列
dfs(board,n-1,i);
}
// System.out.print("[");
for (int i=0;i<n;i++) {
for (int j=0;j<m;j++) {
if (board[i][j] != 'B') {
board[i][j] = 'X';
} else {
board[i][j] = 'O';
}
}
// System.out.print(Arrays.toString(board[i]));
}
// System.out.print("]");
}
private void dfs(char[][] board, int x, int y) {
if (x < 0 || x >= n || y < 0 || y >= m || board[x][y] != 'O') {
// 截止条件，四个方向的值超出矩阵范围或者当前查找元素不是目标替换元素 则直接返回
return;
}
// 修改当前查找元素为'B'
board[x][y] = 'B';
// 递归查找四个方向的元素
for (int[] ints : direction) {
dfs(board,x+ints[0],y+ints[1]);
}
}
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