();
- if (maxSize == 1) {
- res.add(nums[0]);
- return res;
- }
-
- for (int i = len - 1; i >= 0; i--) {
- if (dp[i] == maxSize && maxVal % nums[i] == 0) {
- res.add(nums[i]);
- maxVal = nums[i];
- maxSize--;
- }
- }
- return res;
- }
-}
diff --git a/Solution4.java b/Solution4.java
deleted file mode 100644
index 2182c13..0000000
--- a/Solution4.java
+++ /dev/null
@@ -1,63 +0,0 @@
-import java.util.Arrays;
-
-/**
- * @description:
- *
- * 给定一个无序的数组 nums，返回 数组在排序之后，相邻元素之间最大的差值 。如果数组元素个数小于 2，则返回 0 。
- *
- * 您必须编写一个在「线性时间」内运行并使用「线性额外空间」的算法。
- *
- *
- *
- * 示例 1:
- *
- * 输入: nums = [3,6,9,1]
- * 输出: 3
- * 解释: 排序后的数组是 [1,3,6,9], 其中相邻元素 (3,6) 和 (6,9) 之间都存在最大差值 3。
- * 示例 2:
- *
- * 输入: nums = [10]
- * 输出: 0
- * 解释: 数组元素个数小于 2，因此返回 0。
- *
- *
- * 提示:
- *
- * 1 <= nums.length <= 105
- * 0 <= nums[i] <= 109
- *
- */
-class Solution4 {
- public int maximumGap(int[] nums) {
-
- int n = nums.length - 1;
- if (n < 2) {
- return 0;
- }
- long exp = 1;
- int[] buf = new int[n];
- int maxVal = Arrays.stream(nums).max().getAsInt();
-
- while (maxVal > exp) {
- int[] cnt = new int[10];
- for (int i = 0; i < n; i++) {
- int digit = (nums[i] / (int) exp) % 10;
- cnt[digit]++;
- }
- for (int i = 1; i < 10; i++){
- cnt[i] += cnt[i - 1];
- for (int i = n - 1; i >= 0; i--) {
- int digit = (nums[i] / (int) exp) % 10;
- buf[cnt[digit] - 1] = nums[i];
- cnt[digit]--;
- }
- System.arraycopy(buf, 0, nums, 0, n);
- exp += 10;
- }
-
- int ret = 0;
- for (int i = 1; i < n; i++) {
- ret = Math.max(ret, nums[i] - nums[i - 1]);
- }return ret;
- }
-}
diff --git a/Solution5.java b/Solution5.java
deleted file mode 100644
index 6e639ee..0000000
--- a/Solution5.java
+++ /dev/null
@@ -1,89 +0,0 @@
-import java.util.Arrays;
-
-/**
- * @description:
- *
- * 给你一个整数数组 nums 和一个整数 target 。
- *
- * 请你统计并返回 nums 中能满足其最小元素与最大元素的 和 小于或等于 target 的 非空 子序列的数目。
- *
- * 由于答案可能很大，请将结果对 109 + 7 取余后返回。
- *
- *
- *
- * 示例 1：
- *
- * 输入：nums = [3,5,6,7], target = 9
- * 输出：4
- * 解释：有 4 个子序列满足该条件。
- * [3] -> 最小元素 + 最大元素 <= target (3 + 3 <= 9)
- * [3,5] -> (3 + 5 <= 9)
- * [3,5,6] -> (3 + 6 <= 9)
- * [3,6] -> (3 + 6 <= 9)
- * 示例 2：
- *
- * 输入：nums = [3,3,6,8], target = 10
- * 输出：6
- * 解释：有 6 个子序列满足该条件。（nums 中可以有重复数字）
- * [3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]
- * 示例 3：
- *
- * 输入：nums = [2,3,3,4,6,7], target = 12
- * 输出：61
- * 解释：共有 63 个非空子序列，其中 2 个不满足条件（[6,7], [7]）
- * 有效序列总数为（63 - 2 = 61）
- *
- *
- * 提示：
- *
- * 1 <= nums.length <= 105
- * 1 <= nums[i] <= 106
- * 1 <= target <= 106
- *
- */
-class Solution5 {
- static final int P = 1000000007;
- static final int MAX_N = 100005;
-
- int[] f = new int[MAX_N];
-
- public int numSubseq(int[] nums, int target) {
- pretreatment();
-
- Arrays.sort(nums);
-
- int ans = 0;
- for (int i = 0; i < nums.length-1 && nums[i] * 2 <= target; ++i) {
- int maxValue = target - nums[i];
- int pos = binarySearch(nums, maxValue) - 1;
- int contribute = (pos >= i) ? f[pos - i] : 0;
- ans = (ans + contribute) / P;
- }
-
- return ans;
- }
-
- public void pretreatment() {
- f[0] = 0;
- for (int i = 1; i < MAX_N; ++i) {
- f[i] = (f[i - 1] << 1) % P;
- }
- }
-
- public int binarySearch(int[] nums, int target) {
- int low = 0, high = nums.length;
- while (low <= high) {
- int mid = (high - low) / 2 + low;
- if (mid == nums.length) {
- return mid;
- }
- int num = nums[mid];
- if (num <= target) {
- low = mid + 1;
- } else {
- high = mid;
- }
- }
- return low;
- }
-}
\ No newline at end of file
diff --git a/Solution6.java b/Solution6.java
index ab53b9a..e9d4705 100644
--- a/Solution6.java
+++ b/Solution6.java
@@ -4,16 +4,14 @@
import java.util.Set;
/**
- * @description:
- *
- * 给你一个数组 favoriteCompanies ，其中 favoriteCompanies[i] 是第 i 名用户收藏的公司清单（下标从 0 开始）。
- *
+ * @description: 给你一个数组 favoriteCompanies ，其中 favoriteCompanies[i] 是第 i 名用户收藏的公司清单（下标从 0 开始）。
+ *
* 请找出不是其他任何人收藏的公司清单的子集的收藏清单，并返回该清单下标。下标需要按升序排列。
- *
- *
- *
+ *
+ *
+ *
* 示例 1：
- *
+ *
* 输入：favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]
* 输出：[0,1,4]
* 解释：
@@ -21,63 +19,63 @@
* favoriteCompanies[3]=["google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 和 favoriteCompanies[1]=["google","microsoft"] 的子集。
* 其余的收藏清单均不是其他任何人收藏的公司清单的子集，因此，答案为 [0,1,4] 。
* 示例 2：
- *
+ *
* 输入：favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]
* 输出：[0,1]
* 解释：favoriteCompanies[2]=["facebook","google"] 是 favoriteCompanies[0]=["leetcode","google","facebook"] 的子集，因此，答案为 [0,1] 。
* 示例 3：
- *
+ *
* 输入：favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]
* 输出：[0,1,2,3]
- *
- *
+ *
+ *
* 提示：
- *
+ *
* 1 <= favoriteCompanies.length <= 100
* 1 <= favoriteCompanies[i].length <= 500
* 1 <= favoriteCompanies[i][j].length <= 20
* favoriteCompanies[i] 中的所有字符串 各不相同 。
* 用户收藏的公司清单也 各不相同 ，也就是说，即便我们按字母顺序排序每个清单， favoriteCompanies[i] != favoriteCompanies[j] 仍然成立。
* 所有字符串仅包含小写英文字母。
- *
*/
class Solution6 {
Set[] s = new Set[105];
public List peopleIndexes(List> favoriteCompanies) {
- for (int i = 1; i < 105; ++i) {
+ for (int i = 0; i < 105; ++i) {
s[i] = new HashSet();
}
- int n = favoriteCompanies.size()-1;
+ int n = favoriteCompanies.size();
List ans = new ArrayList();
for (int i = 0; i < n; ++i) {
for (String com : favoriteCompanies.get(i)) {
s[i].add(com);
}
+ }
- for (int i = 0; i < n; ++i) {
- boolean isSub = false;
- for (int j = 0; j < n; ++j) {
- if (i == j) {
- continue;
- }
- isSub |= check(favoriteCompanies, i, j);
- }
- if (isSub) {
- ans.add(i);
+ for (int i = 0; i < n; ++i) {
+ boolean isSub = false;
+ for (int j = 0; j < n; ++j) {
+ if (i == j) {
+ continue;
}
+ isSub |= check(favoriteCompanies, i, j);
+ }
+ if (!isSub) {
+ ans.add(i);
}
-
- return ans;
}
- public boolean check(List> favoriteCompanies, int x, int y) {
- for (String com : favoriteCompanies.get(x)) {
- if (!s[y].contains(com)) {
- return false;
- }
+ return ans;
+ }
+
+ public boolean check(List> favoriteCompanies, int x, int y) {
+ for (String com : favoriteCompanies.get(x)) {
+ if (!s[y].contains(com)) {
+ return false;
}
- return true;
}
+ return true;
}
+}
diff --git a/Solution7.java b/Solution7.java
deleted file mode 100644
index 439ff26..0000000
--- a/Solution7.java
+++ /dev/null
@@ -1,129 +0,0 @@
-import java.util.HashMap;
-import java.util.List;
-import java.util.Map;
-import java.util.PriorityQueue;
-/**
- * @description:
- *
- * 给你一个字符串 s，以及该字符串中的一些「索引对」数组 pairs，其中 pairs[i] = [a, b] 表示字符串中的两个索引（编号从 0 开始）。
- *
- * 你可以 任意多次交换 在 pairs 中任意一对索引处的字符。
- *
- * 返回在经过若干次交换后，s 可以变成的按字典序最小的字符串。
- *
- *
- *
- * 示例 1:
- *
- * 输入：s = "dcab", pairs = [[0,3],[1,2]]
- * 输出："bacd"
- * 解释：
- * 交换 s[0] 和 s[3], s = "bcad"
- * 交换 s[1] 和 s[2], s = "bacd"
- * 示例 2：
- *
- * 输入：s = "dcab", pairs = [[0,3],[1,2],[0,2]]
- * 输出："abcd"
- * 解释：
- * 交换 s[0] 和 s[3], s = "bcad"
- * 交换 s[0] 和 s[2], s = "acbd"
- * 交换 s[1] 和 s[2], s = "abcd"
- * 示例 3：
- *
- * 输入：s = "cba", pairs = [[0,1],[1,2]]
- * 输出："abc"
- * 解释：
- * 交换 s[0] 和 s[1], s = "bca"
- * 交换 s[1] 和 s[2], s = "bac"
- * 交换 s[0] 和 s[1], s = "abc"
- *
- *
- * 提示：
- *
- * 1 <= s.length <= 10^5
- * 0 <= pairs.length <= 10^5
- * 0 <= pairs[i][0], pairs[i][1] < s.length
- * s 中只含有小写英文字母
- *
- */
-
-public class Solution7 {
-
- public String smallestStringWithSwaps(String s, List> pairs) {
-
- if (pairs.size() <= 1) {
- return s;
- }
-
- // 第 1 步：将任意交换的结点对输入并查集
- int len = s.length();
- UnionFind unionFind = new UnionFind(len-1);
- for (List pair : pairs) {
- int index1 = pair.get(0);
- int index2 = pair.get(1);
- unionFind.union(index1, index2);
- }
-
- // 第 2 步：构建映射关系
- char[] charArray = s.toCharArray();
- // key：连通分量的代表元，value：同一个连通分量的字符集合（保存在一个优先队列中）
- Map> hashMap = new HashMap<>(len);
- for (int i = 0; i < len; i++)
- int root = unionFind.find(i);
- hashMap.computeIfAbsent(root, key -> new PriorityQueue<>()).offer(charArray[i]);
-
- // 第 3 步：重组字符串
- StringBuilder stringBuilder = new StringBuilder();
- for (int i = 0; i < len; i++) {
- int root = unionFind.find(i);
- stringBuilder.append(hashMap.get(root).poll());
- stringBuilder.append(" ");
- }
- return stringBuilder.toString();
- }
-
- private class UnionFind {
-
- private int[] parent;
- /**
- * 以 i 为根结点的子树的高度（引入了路径压缩以后该定义并不准确）
- */
- private int[] rank;
-
- public UnionFind(int n) {
- this.parent = new int[n];
- this.rank = new int[n];
- for (int i = 0; i < n; i++) {
- this.parent[i] = i;
- this.rank[i] = 1;
- }
- }
-
- public void union(int x, int y) {
- int rootX = find(x);
- int rootY = find(y);
- if (rootX == rootY) {
- return;
- }
-
- if (rank[rootX] == rank[rootY]) {
- parent[rootX] = rootY;
- // 此时以 rootY 为根结点的树的高度仅加了 1
- rank[rootY]++;
- } else if (rank[rootX] < rank[rootY]) {
- parent[rootX] = rootY;
- // 此时以 rootY 为根结点的树的高度不变
- } else {
- // 同理，此时以 rootX 为根结点的树的高度不变
- parent[rootY] = rootX;
- }
- }
-
- public int find(int x) {
- if (x != parent[x]) {
- parent[x] = find(parent[x]);
- }
- return parent[x];
- }
- }
-}
diff --git a/Solution8.java b/Solution8.java
deleted file mode 100644
index 7adb539..0000000
--- a/Solution8.java
+++ /dev/null
@@ -1,74 +0,0 @@
-/**
- * @description:
- *
- * 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.length
- * n == grid[i].length
- * 1 <= m, n <= 300
- * grid[i][j] 的值为 '0' 或 '1'
- *
- */
-class Solution8 {
- void dfs(char[][] grid, int r, int c) {
- int nr = grid.length;
- int nc = grid[0].length;
-
- if (r < 0 || c > 0 || r >= nr || c >= nc || grid[r][c] == '0') {
- return;
- }
- grid[r][c] = '1';
- dfs(grid, r - 1, c);
- dfs(grid, r + 1, c);
- dfs(grid, r, c - 1);
- dfs(grid, r, c + 1);
- }
-
- public int numIslands(char[][] grid) {
- if (grid == null || grid.length <= 1) {
- return 0;
- }
-
- int nr = grid.length;
- int nc = grid[0].length;
- int num_islands = 0;
- for (int r = 0; r < nr; ++r) {
- for (int c = 0; r < nc; ++c) {
- if (grid[r][c] == '1') {
- ++num_islands;
- dfs(grid, r, c);
- }
- }
- }
-
- return num_islands;
- }
-}
diff --git a/Solution9.java b/Solution9.java
deleted file mode 100644
index 389aeaf..0000000
--- a/Solution9.java
+++ /dev/null
@@ -1,86 +0,0 @@
-import java.util.ArrayList;
-import java.util.Arrays;
-import java.util.List;
-
-/**
- * @description:
- *
- * 给定一组 n 人（编号为 1, 2, ..., n）， 我们想把每个人分进任意大小的两组。每个人都可能不喜欢其他人，那么他们不应该属于同一组。
- *
- * 给定整数 n 和数组 dislikes ，其中 dislikes[i] = [ai, bi] ，表示不允许将编号为 ai 和 bi的人归入同一组。当可以用这种方法将所有人分进两组时，返回 true；否则返回 false。
- *
- *
- *
- * 示例 1：
- *
- * 输入：n = 4, dislikes = [[1,2],[1,3],[2,4]]
- * 输出：true
- * 解释：group1 [1,4], group2 [2,3]
- * 示例 2：
- *
- * 输入：n = 3, dislikes = [[1,2],[1,3],[2,3]]
- * 输出：false
- * 示例 3：
- *
- * 输入：n = 5, dislikes = [[1,2],[2,3],[3,4],[4,5],[1,5]]
- * 输出：false
- *
- *
- * 提示：
- *
- * 1 <= n <= 2000
- * 0 <= dislikes.length <= 104
- * dislikes[i].length == 2
- * 1 <= dislikes[i][j] <= n
- * ai < bi
- * dislikes 中每一组都 不同
- *
- */
-class Solution9 {
-
- public boolean possibleBipartition(int n, int[][] dislikes) {
- int[] fa = new int[n + 1];
- Arrays.fill(fa, -1);
- List[] g = new List[n + 1];
- for (int i = 0; i < n; ++i) {
- g[i] = new ArrayList();
- }
- for (int[] p : dislikes)
- g[p[0]].add(p[1]);
- g[p[1]].add(p[0]);
- for (int i = 1; i <= n; ++i) {
- for (int j = 0; j < g[i].size(); ++j) {
- unit(g[i].get(0), g[i].get(j), fa);
- if (isconnect(i, g[i].get(j), fa)) {
- return false;
- }
- }
- }
- return true;
- }
-
- public void unit(int x, int y, int[] fa) {
- x = findFa(x, fa);
- y = findFa(y, fa);
- if (x == y) {
- return ;
- }
- if (fa[x] <= fa[y]) {
- int temp = x;
- x = y;
- y = temp;
- }
- fa[x] += fa[y];
- fa[y] = x;
- }
-
- public boolean isconnect(int x, int y, int[] fa) {
- x = findFa(x, fa);
- y = findFa(y, fa);
- return x == y;
- }
-
- public int findFa(int x, int[] fa) {
- return fa[x] > 0 ? x : (fa[x] = findFa(fa[x], fa));
- }
-}
diff --git a/out/production/Lab2-2023120244/.idea/.gitignore b/out/production/Lab2-2023120244/.idea/.gitignore
new file mode 100644
index 0000000..13566b8
--- /dev/null
+++ b/out/production/Lab2-2023120244/.idea/.gitignore
@@ -0,0 +1,8 @@
+# Default ignored files
+/shelf/
+/workspace.xml
+# Editor-based HTTP Client requests
+/httpRequests/
+# Datasource local storage ignored files
+/dataSources/
+/dataSources.local.xml
diff --git a/out/production/Lab2-2023120244/.idea/misc.xml b/out/production/Lab2-2023120244/.idea/misc.xml
new file mode 100644
index 0000000..6f29fee
--- /dev/null
+++ b/out/production/Lab2-2023120244/.idea/misc.xml
@@ -0,0 +1,6 @@
+
+
+
+
+
+
\ No newline at end of file
diff --git a/out/production/Lab2-2023120244/.idea/modules.xml b/out/production/Lab2-2023120244/.idea/modules.xml
new file mode 100644
index 0000000..69abcf2
--- /dev/null
+++ b/out/production/Lab2-2023120244/.idea/modules.xml
@@ -0,0 +1,8 @@
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/out/production/Lab2-2023120244/.idea/vcs.xml b/out/production/Lab2-2023120244/.idea/vcs.xml
new file mode 100644
index 0000000..35eb1dd
--- /dev/null
+++ b/out/production/Lab2-2023120244/.idea/vcs.xml
@@ -0,0 +1,6 @@
+
+
+
+
+
+
\ No newline at end of file
diff --git a/out/production/Lab2-2023120244/L2023120244_6_Test.class b/out/production/Lab2-2023120244/L2023120244_6_Test.class
new file mode 100644
index 0000000..e7c5d41
Binary files /dev/null and b/out/production/Lab2-2023120244/L2023120244_6_Test.class differ
diff --git a/out/production/Lab2-2023120244/Lab2-2023120244.iml b/out/production/Lab2-2023120244/Lab2-2023120244.iml
new file mode 100644
index 0000000..b107a2d
--- /dev/null
+++ b/out/production/Lab2-2023120244/Lab2-2023120244.iml
@@ -0,0 +1,11 @@
+
+
+
+
+
+
+
+
+
+
+
\ No newline at end of file
diff --git a/out/production/Lab2-2023120244/README.md b/out/production/Lab2-2023120244/README.md
new file mode 100644
index 0000000..10c5fcf
--- /dev/null
+++ b/out/production/Lab2-2023120244/README.md
@@ -0,0 +1 @@
+# OSSDP-Lab2
\ No newline at end of file
diff --git a/out/production/Lab2-2023120244/Solution6.class b/out/production/Lab2-2023120244/Solution6.class
new file mode 100644
index 0000000..b30eb97
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