给你一个 m x n
的矩阵 board
,由若干字符 'X'
和 'O'
,找到所有被 'X'
围绕的区域,并将这些区域里所有的 'O'
用 'X'
填充。
思路一:深度优先(DFS)
const int dx[4] = {1, -1, 0, 0};
const int dy[4] = {0, 0, 1, -1};
void solve(char** board, int boardSize, int* boardColSize) {
int n = boardSize;
if (n == 0) {
return;
}
int m = boardColSize[0];
int** que = (int**)malloc(sizeof(int*) * n * m);
for (int i = 0; i < n * m; i++) {
que[i] = (int*)malloc(sizeof(int) * 2);
}
int l = 0, r = 0;
for (int i = 0; i < n; i++) {
if (board[i][0] == 'O') {
board[i][0] = 'A';
que[r][0] = i, que[r++][1] = 0;
}
if (board[i][m - 1] == 'O') {
board[i][m - 1] = 'A';
que[r][0] = i, que[r++][1] = m - 1;
}
}
for (int i = 1; i < m - 1; i++) {
if (board[0][i] == 'O') {
board[0][i] = 'A';
que[r][0] = 0, que[r++][1] = i;
}
if (board[n - 1][i] == 'O') {
board[n - 1][i] = 'A';
que[r][0] = n - 1, que[r++][1] = i;
}
}
while (l < r) {
int x = que[l][0], y = que[l][1];
l++;
for (int i = 0; i < 4; i++) {
int mx = x + dx[i], my = y + dy[i];
if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {
continue;
}
board[mx][my] = 'A';
que[r][0] = mx, que[r++][1] = my;
}
}
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (board[i][j] == 'A') {
board[i][j] = 'O';
} else if (board[i][j] == 'O') {
board[i][j] = 'X';
}
}
}
for (int i = 0; i < n * m; i++) {
free(que[i]);
}
free(que);
}
分析:
本题若从内向外分析则难以判断是否需要将O替换为X,所以从外向内分析。从外向内分析时可考虑将最外层的O先替换为A,再判断内部O是否与A相邻,并设置方向数组方便判断方向,最后从外向内递归数组即可完成文章来源:https://www.toymoban.com/news/detail-701145.html
总结:
本题考察对深度优先搜索的应用,分析出从外向内不断判断的方法,再将边界情况考虑清楚即可解决文章来源地址https://www.toymoban.com/news/detail-701145.html
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