目录
三、滑动窗口
30. 长度最小的子数组 ②
31. 无重复字符的最长子串 ②
32. 串联所有单词的子串 ③
33. 最小覆盖子串 ③
四、矩阵
34. 有效的数独 ②
35. 螺旋矩阵 ②
36. 旋转图像 ②
37. 矩阵置零 ②
38. 生命游戏 ②
三、滑动窗口
30. 长度最小的子数组 ②
给定一个含有 n
个正整数的数组和一个正整数 target
。
找出该数组中满足其总和大于等于 target
的长度最小的 连续子数组 [numsl, numsl+1, ..., numsr-1, numsr]
,并返回其长度。如果不存在符合条件的子数组,返回 0
。
示例 1:
输入:target = 7, nums = [2,3,1,2,4,3] 输出:2 解释:子数组 [4,3] 是该条件下的长度最小的子数组。
示例 2:
输入:target = 4, nums = [1,4,4] 输出:1
示例 3:
输入:target = 11, nums = [1,1,1,1,1,1,1,1] 输出:0
提示:
1 <= target <= 109
1 <= nums.length <= 105
1 <= nums[i] <= 105
力扣题解:. - 力扣(LeetCode)
方法1:时间超时(通过15/18)
public int minSubArrayLen(int target, int[] nums) {
int min = nums.length + 1;
for (int i = 0; i < nums.length; i++) {
int sum = 0;
int left = i;
sum += nums[left];
int right = i + 1;
while (right < nums.length && sum < target){
sum += nums[right];
right++;
}
if (sum >= target && min > right - left){
min = right - left;
}
}
return min == nums.length + 1? 0 : min;
}
方法2:(0ms)
public int minSubArrayLen(int target, int[] nums) {
int l = 0, r = 0;
int n = nums.length;
int sum = 0;
while(r < n && sum < target)
sum += nums[r++];
if(r == n)
if (sum < target)
return 0;
else{
while(sum > target)
sum -= nums[l++];
}
while(r < n){
if(sum < target) sum += nums[r++];
sum -= nums[l++];
}
if(sum < target) return r-l+1;
return r -l;
}
方法3:
public int minSubArrayLen(int target, int[] nums) {
int lo = 0, hi = 0, sum = 0, min = Integer.MAX_VALUE;
while (hi < nums.length) {
sum += nums[hi++];
while (sum >= target) {
min = Math.min(min, hi - lo);
sum -= nums[lo++];
}
}
return min == Integer.MAX_VALUE ? 0 : min;
}
方法4:(2ms)
public int minSubArrayLen(int target, int[] nums) {
int left = 0;
int res = Integer.MAX_VALUE;
int add = 0;
for (int right = 0; right < nums.length; right++) {
add += nums[right];
while (add >= target) {
res = Math.min(res, right - left + 1);
add -= nums[left];
++left;
}
}
return res > nums.length ? 0 : res;
}
31. 无重复字符的最长子串 ②
给定一个字符串 s
,请你找出其中不含有重复字符的 最长子串 的长度。
示例 1:
输入: s = "abcabcbb" 输出: 3 解释: 因为无重复字符的最长子串是 "abc",所以其长度为 3。
示例 2:
输入: s = "bbbbb" 输出: 1 解释: 因为无重复字符的最长子串是 "b",所以其长度为 1。
示例 3:
输入: s = "pwwkew" 输出: 3 解释: 因为无重复字符的最长子串是 "wke",所以其长度为 3。 请注意,你的答案必须是 子串 的长度,"pwke" 是一个子序列,不是子串。
提示:
0 <= s.length <= 5 * 104
-
s
由英文字母、数字、符号和空格组成
方法1:(7ms)
public static int lengthOfLongestSubstring(String s) {
TreeSet<Integer> set = new TreeSet<>();
if (s.length() == 0){
return 0;
}else if (s.length() == 1){
return 1;
}else {
int left = 0;
int right = 1;
while (right < s.length()){
String substring = s.substring(left, right);
if (substring.contains(s.charAt(right) + "")){
set.add(right - left);
left = s.indexOf(s.charAt(right), left) + 1;
}else {
if (right == s.length() - 1) {
set.add(right - left + 1);
break;
}
}
right++;
}
return set.last();
}
}
方法2:(滑动窗口 5ms)
public int lengthOfLongestSubstring(String s) {
if (s.length()==0) return 0;
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
int max = 0;
int left = 0;
for(int i = 0; i < s.length(); i ++){
if(map.containsKey(s.charAt(i))){
left = Math.max(left,map.get(s.charAt(i)) + 1);
}
map.put(s.charAt(i),i);
max = Math.max(max,i-left+1);
}
return max;
}
作者:powcai
链接:https://leetcode.cn/problems/longest-substring-without-repeating-characters/solutions/3982/hua-dong-chuang-kou-by-powcai/
32. 串联所有单词的子串 ③
33. 最小覆盖子串 ③
四、矩阵
34. 有效的数独 ②
请你判断一个 9 x 9
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)
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。
- 空白格用
'.'
表示。
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 == 9
board[i].length == 9
-
board[i][j]
是一位数字(1-9
)或者'.'
方法1:
public static boolean isValidSudoku(char[][] board) {
String[] rows = new String[9];
String[] columns = new String[9];
String[] areas = new String[9];
for (int i = 0; i < 9; i++) {
if (rows[i] == null){
rows[i] = "a";
}
for (int j = 0; j < 9; j++) {
if (columns[j] == null){
columns[j] = "a";
}
if (rows[i].contains(board[i][j] + "")){
return false;
}else if (board[i][j] != '.' && !rows[i].contains(board[i][j] + "")){
rows[i] += board[i][j];
}
if (columns[j].contains(board[i][j] + "")){
return false;
}else if (board[i][j] != '.' && !columns[j].contains(board[i][j] + "")){
columns[j] += board[i][j];
}
int row = i / 3; // 0 1 2
int column = j / 3; //0 1 2
int area = row * 3 + column;
if (areas[area] == null){
areas[area] = "a";
}
if (areas[area].contains(board[i][j] + "")){
return false;
}else if (board[i][j] != '.' && !areas[area].contains(board[i][j] + "")){
areas[area] += board[i][j];
}
}
}
return true;
}
方法2:(0ms)
public boolean isValidSudoku(char[][] board) {
short[] rows = new short[9];
short[] cols = new short[9];
short[] squares = new short[9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
short value = (short)(1 << (board[i][j] - '0'));
int boxNum = i / 3 * 3 + j / 3;
if ((rows[i] & value) != 0 || (cols[j] & value) != 0 || (squares[boxNum] & value) != 0) {
return false;
}
rows[i] |= value;
cols[j] |= value;
squares[boxNum] |= value;
}
}
}
return true;
}
方法3:(1ms)
public boolean isValidSudoku(char[][] board) {
int[] rows = new int[9];
int[] cols = new int[9];
int[] subboxes = new int[9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char c = board[i][j];
if (c != '.') {
int x = (1 << (c - '1'));
if ((rows[i] & x) != 0) {
return false;
} else {
rows[i] |= x;
}
if ((cols[j] & x) != 0) {
return false;
} else {
cols[j] |= x;
}
if ((subboxes[i / 3 * 3 + j / 3] & x) != 0) {
return false;
} else {
subboxes[i / 3 * 3 + j / 3] |= x;
}
}
}
}
return true;
}
方法4:(2ms)
public boolean isValidSudoku(char[][] board) {
int[][] row = new int[9][9]; // 行
int[][] columns = new int[9][9]; // 列
int[][][] a = new int[3][3][9];
for (int i = 0; i < 9; i++) {
Set<Character> set = new HashSet<>();
for (int j = 0; j < 9; j++) {
char c = board[i][j];
if (c != '.') {
row[i][c - 49]++;
columns[j][c - 49]++;
a[i / 3][j / 3][c - 49]++;
if (row[i][c - 49] > 1 || columns[j][c - 49] > 1 || a[i / 3][j / 3][c - 49] > 1) {
return false;
}
}
}
}
return true;
}
35. 螺旋矩阵 ②
给你一个 m
行 n
列的矩阵 matrix
,请按照 顺时针螺旋顺序 ,返回矩阵中的所有元素。
示例 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[1,2,3,6,9,8,7,4,5]
示例 2:
输入:matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]] 输出:[1,2,3,4,8,12,11,10,9,5,6,7]
提示:
m == matrix.length
n == matrix[i].length
1 <= m, n <= 10
-100 <= matrix[i][j] <= 100
方法1:(0ms)
public static List<Integer> spiralOrder(int[][] matrix) {
int[][] path = new int[matrix.length][matrix[0].length];
int step = 0;
ArrayList<Integer> list = new ArrayList<>();
int row = 0;
int col = 0;
int direct = 0;
while (step < matrix.length * matrix[0].length){
list.add(matrix[row][col]);
path[row][col] = 1;
if (direct == 0){
col++;
if (col == matrix[0].length || path[row][col] == 1){
direct = 1;
col--;
row++;
}
}else if (direct == 1){
row++;
if (row == matrix.length || path[row][col] == 1){
direct = 2;
row--;
col--;
}
}else if (direct == 2){
col--;
if (col == -1 || path[row][col] == 1){
direct = 3;
col++;
row--;
}
}else {
row--;
if (row == 0 || path[row][col] == 1){
direct = 0;
row++;
col++;
}
}
step++;
}
return list;
}
方法2:(0ms)
public List<Integer> spiralOrder(int[][] matrix) {
int m = matrix.length, n = matrix[0].length;
List<Integer> res = new ArrayList<>();
int u = 0, d = m - 1, l = 0, r = n - 1;
while (true) {
for (int i = l; i <= r; i ++) res.add(matrix[u][i]);
if (++u > d) break;
for (int i = u; i <= d; i ++) res.add(matrix[i][r]);
if (--r < l) break;
for (int i = r; i >= l; i --) res.add(matrix[d][i]);
if (--d < u) break;
for (int i = d; i >= u; i --) res.add(matrix[i][l]);
if (++l > r) break;
}
return res;
}
36. 旋转图像 ②
给定一个 n × n 的二维矩阵 matrix
表示一个图像。请你将图像顺时针旋转 90 度。
你必须在 原地 旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。
示例 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]] 输出:[[7,4,1],[8,5,2],[9,6,3]]
示例 2:
输入:matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]] 输出:[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
提示:
n == matrix.length == matrix[i].length
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000
方法1:(0ms)
public static void rotate(int[][] matrix) {
int n = matrix.length;
int row = 0;
int col = n - 1;
int index = 0;
while (row < col){
while (row + index < col) {
int leftUp = matrix[row][row + index];
int rightUp = matrix[row + index][col];
int rightDown = matrix[col][col - index];
int leftDown = matrix[col - index][row];
matrix[row][row + index] = leftDown;
matrix[row + index][col] = leftUp;
matrix[col][col - index] = rightUp;
matrix[col - index][row] = rightDown;
index++;
}
index = 0;
row++;
col--;
}
}
37. 矩阵置零 ②
给定一个 m x n
的矩阵,如果一个元素为 0 ,则将其所在行和列的所有元素都设为 0 。请使用 原地 算法。
示例 1:
输入:matrix = [[1,1,1],[1,0,1],[1,1,1]] 输出:[[1,0,1],[0,0,0],[1,0,1]]
示例 2:
输入:matrix = [[0,1,2,0],[3,4,5,2],[1,3,1,5]] 输出:[[0,0,0,0],[0,4,5,0],[0,3,1,0]]
提示:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
进阶:文章来源:https://www.toymoban.com/news/detail-851160.html
- 一个直观的解决方案是使用
O(mn)
的额外空间,但这并不是一个好的解决方案。 - 一个简单的改进方案是使用
O(m + n)
的额外空间,但这仍然不是最好的解决方案。 - 你能想出一个仅使用常量空间的解决方案吗?
方法1:
public void setZeroes(int[][] matrix) {
TreeSet<Integer> rowSet = new TreeSet<>();
TreeSet<Integer> columnSet = new TreeSet<>();
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 0){
rowSet.add(i);
columnSet.add(j);
}
}
}
for (int i = 0; i < matrix.length; i++) {
if (rowSet.contains(i)){
Arrays.fill(matrix[i], 0);
}
}
for (int i = 0; i < matrix[0].length; i++) {
if (columnSet.contains(i)){
for (int j = 0; j < matrix.length; j++) {
matrix[j][i] = 0;
}
}
}
}
方法2:
public void setZeroes(int[][] matrix) {
int m = matrix.length;
int n = matrix[0].length;
// 统计行列是否需要置为0 空间复杂度 O(m+n)
boolean[] zeroRow = new boolean[m];
boolean[] zeroCol = new boolean[n];
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(matrix[i][j] == 0){
zeroRow[i] = true;
zeroCol[j] = true;
}
}
}
for(int i = 0; i < m; i++) {
for(int j = 0; j < n; j++) {
if(zeroRow[i] || zeroCol[j]) {
matrix[i][j] = 0;
}
}
}
}
38. 生命游戏 ②
根据 百度百科 , 生命游戏 ,简称为 生命 ,是英国数学家约翰·何顿·康威在 1970 年发明的细胞自动机。
给定一个包含 m × n
个格子的面板,每一个格子都可以看成是一个细胞。每个细胞都具有一个初始状态: 1
即为 活细胞 (live),或 0
即为 死细胞 (dead)。每个细胞与其八个相邻位置(水平,垂直,对角线)的细胞都遵循以下四条生存定律:
- 如果活细胞周围八个位置的活细胞数少于两个,则该位置活细胞死亡;
- 如果活细胞周围八个位置有两个或三个活细胞,则该位置活细胞仍然存活;
- 如果活细胞周围八个位置有超过三个活细胞,则该位置活细胞死亡;
- 如果死细胞周围正好有三个活细胞,则该位置死细胞复活;
下一个状态是通过将上述规则同时应用于当前状态下的每个细胞所形成的,其中细胞的出生和死亡是同时发生的。给你 m x n
网格面板 board
的当前状态,返回下一个状态。
示例 1:
输入:board = [[0,1,0],[0,0,1],[1,1,1],[0,0,0]] 输出:[[0,0,0],[1,0,1],[0,1,1],[0,1,0]]
示例 2:
输入:board = [[1,1],[1,0]] 输出:[[1,1],[1,1]]
提示:
m == board.length
n == board[i].length
1 <= m, n <= 25
-
board[i][j]
为0
或1
进阶:
- 你可以使用原地算法解决本题吗?请注意,面板上所有格子需要同时被更新:你不能先更新某些格子,然后使用它们的更新后的值再更新其他格子。
- 本题中,我们使用二维数组来表示面板。原则上,面板是无限的,但当活细胞侵占了面板边界时会造成问题。你将如何解决这些问题?
方法1:(0ms)文章来源地址https://www.toymoban.com/news/detail-851160.html
public static void gameOfLife(int[][] board) {
int[][] map = new int[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
int row = Math.max(i - 1, 0);
int col = Math.max(j - 1, 0);
int src = board[i][j];
int count = 0;
while (row < board.length && col < board[0].length){
count += board[row][col];
col++;
if (col == j + 2 || col == board[0].length){
row++;
col = Math.max(j - 1, 0);
}
if (row == i + 2 || row == board.length){
break;
}
}
count -= src;
if (src == 1 && count == 1) {
map[i][j] = 0;
}else if (src == 1 && (count == 2 || count == 3)){
map[i][j] = 1;
}else if (src == 1 && count > 3){
map[i][j] = 0;
}else if (src == 0 && count == 3){
map[i][j] = 1;
}
}
}
// board = map;
// for (int[] ints : board) {
// System.out.println(Arrays.toString(ints));
// }
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
board[i][j] = map[i][j];
}
}
}
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