0.简介
1.图像渲染
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
int[] dx = { 0, 0, 1, -1 };
int[] dy = { 1, -1, 0, 0 };
int m, n;
int prev;
public int[][] floodFill(int[][] image, int sr, int sc, int color) {
if (image[sr][sc] == color)
return image;
m = image.length;
n = image[0].length;
prev = image[sr][sc];
dfs(image, sr, sc, color);
return image;
}
public void dfs(int[][] image, int i, int j, int color) {
image[i][j] = color;
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && image[x][y] == prev) {
dfs(image, x, y, color);
}
}
}
}2.岛屿数量
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
boolean[][] vis;
int m, n;
int[] dx = { 0, 0, -1, 1 };
int[] dy = { 1, -1, 0, 0 };
public int numIslands(char[][] grid) {
m = grid.length;
n = grid[0].length;
vis = new boolean[m][n];
int ret = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (!vis[i][j] && grid[i][j] == '1') {
ret++;
dfs(grid, i, j);
}
}
return ret;
}
public void dfs(char[][] grid, int i, int j) {
vis[i][j] = true;
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == '1') {
dfs(grid, x, y);
}
}
}
}3.岛屿的最大面积
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
boolean[][] vis;
int m, n;
int[] dx = { 0, 0, -1, 1 };
int[] dy = { 1, -1, 0, 0 };
int count;
public int maxAreaOfIsland(int[][] grid) {
m = grid.length;
n = grid[0].length;
vis = new boolean[m][n];
int ret = 0;
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (!vis[i][j] && grid[i][j] == 1) {
count = 0;
dfs(grid, i, j);
ret = Math.max(ret, count);
}
}
return ret;
}
public void dfs(int[][] grid, int i, int j) {
count++;
vis[i][j] = true;
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1)
dfs(grid, x, y);
}
}
}4.被围绕的区域
130. 被围绕的区域 - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
int m, n;
int[] dx = { 1, -1, 0, 0 };
int[] dy = { 0, 0, 1, -1 };
public void solve(char[][] board) {
m = board.length;
n = board[0].length;
// 1. 先把边界的 O 相连的联通块，全部修改成 '.'
for (int j = 0; j < n; j++) {
if (board[0][j] == 'O')
dfs(board, 0, j);
if (board[m - 1][j] == 'O')
dfs(board, m - 1, j);
}
for (int i = 0; i < m; i++) {
if (board[i][0] == 'O')
dfs(board, i, 0);
if (board[i][n - 1] == 'O')
dfs(board, i, n - 1);
}
// 2. 还原
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++) {
if (board[i][j] == '.')
board[i][j] = 'O';
else if (board[i][j] == 'O')
board[i][j] = 'X';
}
}
public void dfs(char[][] board, int i, int j) {
board[i][j] = '.';
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O') {
dfs(board, x, y);
}
}
}
}5.太平洋大西洋水流问题
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
int m, n;
int[] dx = { 0, 0, 1, -1 };
int[] dy = { 1, -1, 0, 0 };
public List<List<Integer>> pacificAtlantic(int[][] h) {
m = h.length;
n = h[0].length;
boolean[][] pac = new boolean[m][n];
boolean[][] atl = new boolean[m][n];
// 1. 先处理 pac 洋
for (int j = 0; j < n; j++)
dfs(h, 0, j, pac);
for (int i = 0; i < m; i++)
dfs(h, i, 0, pac);
// 2. 再处理 atl 洋
for (int i = 0; i < m; i++)
dfs(h, i, n - 1, atl);
for (int j = 0; j < n; j++)
dfs(h, m - 1, j, atl);
// 3. 提取结果
List<List<Integer>> ret = new ArrayList<>();
for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (pac[i][j] && atl[i][j]) {
List<Integer> tmp = new ArrayList<>();
tmp.add(i);
tmp.add(j);
ret.add(tmp);
}
return ret;
}
public void dfs(int[][] h, int i, int j, boolean[][] vis) {
vis[i][j] = true;
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && h[x][y] >= h[i][j]) {
dfs(h, x, y, vis);
}
}
}
}6.扫雷问题
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
int[] dx = { 0, 0, 1, -1, 1, 1, -1, -1 };
int[] dy = { 1, -1, 0, 0, 1, -1, 1, -1 };
int m, n;
public char[][] updateBoard(char[][] board, int[] click) {
m = board.length;
n = board[0].length;
int x = click[0], y = click[1];
if (board[x][y] == 'M') // 如果直接点到地雷，游戏结束
{
board[x][y] = 'X';
return board;
}
dfs(board, x, y);
return board;
}
public void dfs(char[][] board, int i, int j) {
// 统计⼀下周围地雷的个数
int count = 0;
for (int k = 0; k < 8; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
count++;
}
}
if (count == 0) // 周围没有地雷
{
board[i][j] = 'B';
for (int k = 0; k < 8; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
dfs(board, x, y);
}
}
} else {
board[i][j] = (char) ('0' + count);
return;
}
}
}7.机器人的运动范围
. - 力扣（LeetCode）
题目解析
算法原理
代码
class Solution {
int m, n, k;
boolean[][] vis;
int[] dx = { 1, -1, 0, 0 };
int[] dy = { 0, 0, 1, -1 };
int ret;
public int movingCount(int _m, int _n, int _k) {
m = _m;
n = _n;
k = _k;
vis = new boolean[m][n];
dfs(0, 0);
return ret;
}
public void dfs(int i, int j) {
ret++;
vis[i][j] = true;
for (int k = 0; k < 4; k++) {
int x = i + dx[k], y = j + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && check(x, y)) {
dfs(x, y);
}
}
}
public boolean check(int i, int j) {
int tmp = 0;
while (i != 0) {
tmp += i % 10;
i /= 10;
}
while (j != 0) {
tmp += j % 10;
j /= 10;
}
return tmp <= k;
}
}
