目录
1. 全排列 II 🌟🌟
2. 岛屿数量 🌟🌟
3. 有效的数独 🌟🌟
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1. 全排列 II
给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
示例 1：
输入：nums = [1,1,2] 输出：[[1,1,2], [1,2,1], [2,1,1]]
示例 2：
输入：nums = [1,2,3] 输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
提示：
1 <= nums.length <= 8-10 <= nums[i] <= 10
出处：
https://edu.csdn.net/practice/27137143
代码1：
class Solution {
List<List<Integer>> ans = new ArrayList<>();
public List<List<Integer>> permuteUnique(int[] nums) {
dfs(nums, 0);
return ans;
}
private void dfs(int[] nums, int cur) {
if (cur == nums.length) {
List<Integer> line = new ArrayList<>();
for (int i : nums) {
line.add(i);
}
ans.add(line);
} else {
for (int i = cur; i < nums.length; i++) {
if (canSwap(nums, cur, i)) {
swap(nums, cur, i);
dfs(nums, cur + 1);
swap(nums, cur, i);
}
}
}
}
private boolean canSwap(int nums[], int begin, int end) {
for (int i = begin; i < end; i++) {
if (nums[i] == nums[end]) {
return false;
}
}
return true;
}
private void swap(int nums[], int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}代码2：
import java.util.*;
class permuteUnique {
public static class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
while (true) {
res.add(getList(nums));
if (!nextPermutation(nums)) {
break;
}
}
return res;
}
private List<Integer> getList(int[] nums) {
List<Integer> list = new ArrayList<>();
for (int num : nums) {
list.add(num);
}
return list;
}
private boolean nextPermutation(int[] nums) {
int i = nums.length - 2;
while (i >= 0 && nums[i] >= nums[i + 1]) {
i--;
}
if (i < 0) {
return false;
}
int j = nums.length - 1;
while (j > i && nums[j] <= nums[i]) {
j--;
}
swap(nums, i, j);
reverse(nums, i + 1);
return true;
}
private void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
private void reverse(int[] nums, int start) {
int i = start, j = nums.length - 1;
while (i < j) {
swap(nums, i, j);
i++;
j--;
}
}
}
public static void main(String[] args) {
Solution s = new Solution();
int[] nums = {1,1,2};
System.out.println(s.permuteUnique(nums));
int[] nums2 = {1,2,3};
System.out.println(s.permuteUnique(nums2));
}
}
输出：
[[1, 1, 2], [1, 2, 1], [2, 1, 1]]
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
2. 岛屿数量
给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
此外，你可以假设该网格的四条边均被水包围。
示例 1：
输入：grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ] 输出：1
示例 2：
输入：grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ] 输出：3
提示：
m == grid.lengthn == grid[i].length1 <= m, n <= 300grid[i][j]的值为'0'或'1'
出处：
https://edu.csdn.net/practice/27137144
代码1： dfs
import java.util.*;
class numIslands {
public static class Solution {
public int numIslands(char[][] grid) {
int islandNum = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == '1') {
infect(grid, i, j);
islandNum++;
}
}
}
return islandNum;
}
public void infect(char[][] grid, int i, int j) {
if (i < 0 || i >= grid.length ||
j < 0 || j >= grid[0].length || grid[i][j] != '1') {
return;
}
grid[i][j] = '2';
infect(grid, i + 1, j);
infect(grid, i - 1, j);
infect(grid, i, j + 1);
infect(grid, i, j - 1);
}
}
public static void main(String[] args) {
Solution s = new Solution();
char[][] grid = {
{'1','1','1','1','0'},
{'1','1','0','1','0'},
{'1','1','0','0','0'},
{'0','0','0','0','0'}};
System.out.println(s.numIslands(grid));
char[][] grid2 = {
{'1','1','0','0','0'},
{'1','1','0','0','0'},
{'0','0','1','0','0'},
{'0','0','0','1','1'}};
System.out.println(s.numIslands(grid2));
}
}
代码2： bfs
import java.util.*;
class numIslands {
public static class Solution {
public int numIslands(char[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int islandNum = 0;
int[] dx = {0, 0, -1, 1};
int[] dy = {-1, 1, 0, 0};
Queue<int[]> queue = new LinkedList<>();
boolean[][] visited = new boolean[m][n];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == '1' && !visited[i][j]) {
islandNum++;
visited[i][j] = true;
queue.offer(new int[]{i, j});
while (!queue.isEmpty()) {
int[] curr = queue.poll();
for (int k = 0; k < 4; k++) {
int x = curr[0] + dx[k];
int y = curr[1] + dy[k];
if (x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' && !visited[x][y]) {
visited[x][y] = true;
queue.offer(new int[]{x, y});
}
}
}
}
}
}
return islandNum;
}
}
public static void main(String[] args) {
Solution s = new Solution();
char[][] grid = {
{'1','1','1','1','0'},
{'1','1','0','1','0'},
{'1','1','0','0','0'},
{'0','0','0','0','0'}};
System.out.println(s.numIslands(grid));
char[][] grid2 = {
{'1','1','0','0','0'},
{'1','1','0','0','0'},
{'0','0','1','0','0'},
{'0','0','0','1','1'}};
System.out.println(s.numIslands(grid2));
}
}
输出：
1
3
3. 有效的数独
请你判断一个 9x9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。
- 数字
1-9在每一行只能出现一次。 - 数字
1-9在每一列只能出现一次。 - 数字
1-9在每一个以粗实线分隔的3x3宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
注意：
- 一个有效的数独（部分已被填充）不一定是可解的。
- 只需要根据以上规则，验证已经填入的数字是否有效即可。
示例 1：
输入：board = [["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"]] 输出：true
示例 2：
输入：board = [["8","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"]] 输出：false 解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
提示：
board.length == 9board[i].length == 9board[i][j]是一位数字或者'.'
出处：
https://edu.csdn.net/practice/27137145
代码1：
import java.util.*;
class isValidSudoku {
public static class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] row = new boolean[9][9];
boolean[][] col = new boolean[9][9];
boolean[][] block = new boolean[9][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
int num = board[i][j] - '1';
int blockIndex = i / 3 * 3 + j / 3;
if (row[i][num] || col[j][num] || block[blockIndex][num]) {
return false;
} else {
row[i][num] = true;
col[j][num] = true;
block[blockIndex][num] = true;
}
}
}
}
return true;
}
}
public static void main(String[] args) {
Solution s = new Solution();
char[][] board = {
{'5','3','.','.','7','.','.','.','.'},
{'6','.','.','1','9','5','.','.','.'},
{'.','9','8','.','.','.','.','6','.'},
{'8','.','.','.','6','.','.','.','3'},
{'4','.','.','8','.','3','.','.','1'},
{'7','.','.','.','2','.','.','.','6'},
{'.','6','.','.','.','.','2','8','.'},
{'.','.','.','4','1','9','.','.','5'},
{'.','.','.','.','8','.','.','7','9'}};
System.out.println(s.isValidSudoku(board));
char[][] board2 = {
{'8','3','.','.','7','.','.','.','.'},
{'6','.','.','1','9','5','.','.','.'},
{'.','9','8','.','.','.','.','6','.'},
{'8','.','.','.','6','.','.','.','3'},
{'4','.','.','8','.','3','.','.','1'},
{'7','.','.','.','2','.','.','.','6'},
{'.','6','.','.','.','.','2','8','.'},
{'.','.','.','4','1','9','.','.','5'},
{'.','.','.','.','8','.','.','7','9'}};
System.out.println(s.isValidSudoku(board2));
}
}
代码2：
import java.util.*;
class isValidSudoku {
public static class Solution {
public boolean isValidSudoku(char[][] board) {
Map<Integer, Set<Character>> rows = new HashMap<>();
Map<Integer, Set<Character>> cols = new HashMap<>();
Map<Integer, Set<Character>> boxes = new HashMap<>();
for (int i = 0; i < 9; i++) {
rows.put(i, new HashSet<>());
cols.put(i, new HashSet<>());
boxes.put(i, new HashSet<>());
}
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char num = board[i][j];
if (num == '.') {
continue;
}
if (rows.get(i).contains(num) || cols.get(j).contains(num) || boxes.get((i / 3) * 3 + j / 3).contains(num)) {
return false;
}
rows.get(i).add(num);
cols.get(j).add(num);
boxes.get((i / 3) * 3 + j / 3).add(num);
}
}
return true;
}
}
public static void main(String[] args) {
Solution s = new Solution();
char[][] board = {
{'5','3','.','.','7','.','.','.','.'},
{'6','.','.','1','9','5','.','.','.'},
{'.','9','8','.','.','.','.','6','.'},
{'8','.','.','.','6','.','.','.','3'},
{'4','.','.','8','.','3','.','.','1'},
{'7','.','.','.','2','.','.','.','6'},
{'.','6','.','.','.','.','2','8','.'},
{'.','.','.','4','1','9','.','.','5'},
{'.','.','.','.','8','.','.','7','9'}};
System.out.println(s.isValidSudoku(board));
char[][] board2 = {
{'8','3','.','.','7','.','.','.','.'},
{'6','.','.','1','9','5','.','.','.'},
{'.','9','8','.','.','.','.','6','.'},
{'8','.','.','.','6','.','.','.','3'},
{'4','.','.','8','.','3','.','.','1'},
{'7','.','.','.','2','.','.','.','6'},
{'.','6','.','.','.','.','2','8','.'},
{'.','.','.','4','1','9','.','.','5'},
{'.','.','.','.','8','.','.','7','9'}};
System.out.println(s.isValidSudoku(board2));
}
}
输出：
true
false
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