From 80a78b10a380ac24b595b9a7f69ff818f144591f Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Wed, 5 Feb 2025 07:55:03 +0800
Subject: [PATCH 1/3] feat: add solutions to lc problem: No.0090
No.0090.Subsets II
---
solution/0000-0099/0090.Subsets II/README.md | 129 +++++++++++++++++-
.../0000-0099/0090.Subsets II/README_EN.md | 129 +++++++++++++++++-
.../0000-0099/0090.Subsets II/Solution.cpp | 6 +-
.../0000-0099/0090.Subsets II/Solution.cs | 26 ++++
.../0000-0099/0090.Subsets II/Solution.go | 4 +-
.../0000-0099/0090.Subsets II/Solution.js | 25 ++++
.../0000-0099/0090.Subsets II/Solution2.cpp | 4 +-
.../0000-0099/0090.Subsets II/Solution2.cs | 24 ++++
.../0000-0099/0090.Subsets II/Solution2.js | 26 ++++
9 files changed, 358 insertions(+), 15 deletions(-)
create mode 100644 solution/0000-0099/0090.Subsets II/Solution.cs
create mode 100644 solution/0000-0099/0090.Subsets II/Solution.js
create mode 100644 solution/0000-0099/0090.Subsets II/Solution2.cs
create mode 100644 solution/0000-0099/0090.Subsets II/Solution2.js
diff --git a/solution/0000-0099/0090.Subsets II/README.md b/solution/0000-0099/0090.Subsets II/README.md
index 92f9bc0881acb..a1e8455409916 100644
--- a/solution/0000-0099/0090.Subsets II/README.md
+++ b/solution/0000-0099/0090.Subsets II/README.md
@@ -133,11 +133,11 @@ class Solution {
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
vector<vector<int>> ans;
vector<int> t;
int n = nums.size();
- function<void(int)> dfs = [&](int i) {
+ auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
return;
@@ -160,7 +160,7 @@ public:
```go
func subsetsWithDup(nums []int) (ans [][]int) {
- sort.Ints(nums)
+ slices.Sort(nums)
n := len(nums)
t := []int{}
var dfs func(int)
@@ -239,6 +239,67 @@ impl Solution {
}
```
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const t = [];
+ const ans = [];
+ const dfs = i => {
+ if (i >= n) {
+ ans.push([...t]);
+ return;
+ }
+ t.push(nums[i]);
+ dfs(i + 1);
+ t.pop();
+ while (i + 1 < n && nums[i] === nums[i + 1]) {
+ i++;
+ }
+ dfs(i + 1);
+ };
+ dfs(0);
+ return ans;
+};
+```
+
+#### C#
+
+```cs
+public class Solution {
+ private IList<IList<int>> ans = new List<IList<int>>();
+ private IList<int> t = new List<int>();
+ private int[] nums;
+
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ this.nums = nums;
+ Dfs(0);
+ return ans;
+ }
+
+ private void Dfs(int i) {
+ if (i >= nums.Length) {
+ ans.Add(new List<int>(t));
+ return;
+ }
+ t.Add(nums[i]);
+ Dfs(i + 1);
+ t.RemoveAt(t.Count - 1);
+ while (i + 1 < nums.Length && nums[i + 1] == nums[i]) {
+ ++i;
+ }
+ Dfs(i + 1);
+ }
+}
+```
+
<!-- tabs:end -->
<!-- solution:end -->
@@ -314,7 +375,7 @@ class Solution {
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
int n = nums.size();
vector<vector<int>> ans;
for (int mask = 0; mask < 1 << n; ++mask) {
@@ -421,6 +482,66 @@ impl Solution {
}
```
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const ans = [];
+ for (let mask = 0; mask < 1 << n; ++mask) {
+ const t = [];
+ let ok = true;
+ for (let i = 0; i < n; ++i) {
+ if (((mask >> i) & 1) === 1) {
+ if (i && ((mask >> (i - 1)) & 1) === 0 && nums[i] === nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.push(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.push(t);
+ }
+ }
+ return ans;
+};
+```
+
+#### C#
+
+```cs
+public class Solution {
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ int n = nums.Length;
+ IList<IList<int>> ans = new List<IList<int>>();
+ for (int mask = 0; mask < 1 << n; ++mask) {
+ IList<int> t = new List<int>();
+ bool ok = true;
+ for (int i = 0; i < n; ++i) {
+ if ((mask >> i & 1) == 1) {
+ if (i > 0 && (mask >> (i - 1) & 1) == 0 && nums[i] == nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.Add(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.Add(t);
+ }
+ }
+ return ans;
+ }
+}
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/0000-0099/0090.Subsets II/README_EN.md b/solution/0000-0099/0090.Subsets II/README_EN.md
index 23b33b1caaa0d..cf0c5634e19a1 100644
--- a/solution/0000-0099/0090.Subsets II/README_EN.md
+++ b/solution/0000-0099/0090.Subsets II/README_EN.md
@@ -120,11 +120,11 @@ class Solution {
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
vector<vector<int>> ans;
vector<int> t;
int n = nums.size();
- function<void(int)> dfs = [&](int i) {
+ auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
return;
@@ -147,7 +147,7 @@ public:
```go
func subsetsWithDup(nums []int) (ans [][]int) {
- sort.Ints(nums)
+ slices.Sort(nums)
n := len(nums)
t := []int{}
var dfs func(int)
@@ -226,6 +226,67 @@ impl Solution {
}
```
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const t = [];
+ const ans = [];
+ const dfs = i => {
+ if (i >= n) {
+ ans.push([...t]);
+ return;
+ }
+ t.push(nums[i]);
+ dfs(i + 1);
+ t.pop();
+ while (i + 1 < n && nums[i] === nums[i + 1]) {
+ i++;
+ }
+ dfs(i + 1);
+ };
+ dfs(0);
+ return ans;
+};
+```
+
+#### C#
+
+```cs
+public class Solution {
+ private IList<IList<int>> ans = new List<IList<int>>();
+ private IList<int> t = new List<int>();
+ private int[] nums;
+
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ this.nums = nums;
+ Dfs(0);
+ return ans;
+ }
+
+ private void Dfs(int i) {
+ if (i >= nums.Length) {
+ ans.Add(new List<int>(t));
+ return;
+ }
+ t.Add(nums[i]);
+ Dfs(i + 1);
+ t.RemoveAt(t.Count - 1);
+ while (i + 1 < nums.Length && nums[i + 1] == nums[i]) {
+ ++i;
+ }
+ Dfs(i + 1);
+ }
+}
+```
+
<!-- tabs:end -->
<!-- solution:end -->
@@ -301,7 +362,7 @@ class Solution {
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
int n = nums.size();
vector<vector<int>> ans;
for (int mask = 0; mask < 1 << n; ++mask) {
@@ -408,6 +469,66 @@ impl Solution {
}
```
+#### JavaScript
+
+```js
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const ans = [];
+ for (let mask = 0; mask < 1 << n; ++mask) {
+ const t = [];
+ let ok = true;
+ for (let i = 0; i < n; ++i) {
+ if (((mask >> i) & 1) === 1) {
+ if (i && ((mask >> (i - 1)) & 1) === 0 && nums[i] === nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.push(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.push(t);
+ }
+ }
+ return ans;
+};
+```
+
+#### C#
+
+```cs
+public class Solution {
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ int n = nums.Length;
+ IList<IList<int>> ans = new List<IList<int>>();
+ for (int mask = 0; mask < 1 << n; ++mask) {
+ IList<int> t = new List<int>();
+ bool ok = true;
+ for (int i = 0; i < n; ++i) {
+ if ((mask >> i & 1) == 1) {
+ if (i > 0 && (mask >> (i - 1) & 1) == 0 && nums[i] == nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.Add(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.Add(t);
+ }
+ }
+ return ans;
+ }
+}
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/0000-0099/0090.Subsets II/Solution.cpp b/solution/0000-0099/0090.Subsets II/Solution.cpp
index 49b08baab6a7d..488fdd4c07639 100644
--- a/solution/0000-0099/0090.Subsets II/Solution.cpp
+++ b/solution/0000-0099/0090.Subsets II/Solution.cpp
@@ -1,11 +1,11 @@
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
vector<vector<int>> ans;
vector<int> t;
int n = nums.size();
- function<void(int)> dfs = [&](int i) {
+ auto dfs = [&](this auto&& dfs, int i) {
if (i >= n) {
ans.push_back(t);
return;
@@ -21,4 +21,4 @@ class Solution {
dfs(0);
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/0000-0099/0090.Subsets II/Solution.cs b/solution/0000-0099/0090.Subsets II/Solution.cs
new file mode 100644
index 0000000000000..9446f1701ef36
--- /dev/null
+++ b/solution/0000-0099/0090.Subsets II/Solution.cs
@@ -0,0 +1,26 @@
+public class Solution {
+ private IList<IList<int>> ans = new List<IList<int>>();
+ private IList<int> t = new List<int>();
+ private int[] nums;
+
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ this.nums = nums;
+ Dfs(0);
+ return ans;
+ }
+
+ private void Dfs(int i) {
+ if (i >= nums.Length) {
+ ans.Add(new List<int>(t));
+ return;
+ }
+ t.Add(nums[i]);
+ Dfs(i + 1);
+ t.RemoveAt(t.Count - 1);
+ while (i + 1 < nums.Length && nums[i + 1] == nums[i]) {
+ ++i;
+ }
+ Dfs(i + 1);
+ }
+}
diff --git a/solution/0000-0099/0090.Subsets II/Solution.go b/solution/0000-0099/0090.Subsets II/Solution.go
index 70194ede97453..c716f94b56751 100644
--- a/solution/0000-0099/0090.Subsets II/Solution.go
+++ b/solution/0000-0099/0090.Subsets II/Solution.go
@@ -1,5 +1,5 @@
func subsetsWithDup(nums []int) (ans [][]int) {
- sort.Ints(nums)
+ slices.Sort(nums)
n := len(nums)
t := []int{}
var dfs func(int)
@@ -18,4 +18,4 @@ func subsetsWithDup(nums []int) (ans [][]int) {
}
dfs(0)
return
-}
\ No newline at end of file
+}
diff --git a/solution/0000-0099/0090.Subsets II/Solution.js b/solution/0000-0099/0090.Subsets II/Solution.js
new file mode 100644
index 0000000000000..e6b6ae15f625e
--- /dev/null
+++ b/solution/0000-0099/0090.Subsets II/Solution.js
@@ -0,0 +1,25 @@
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const t = [];
+ const ans = [];
+ const dfs = i => {
+ if (i >= n) {
+ ans.push([...t]);
+ return;
+ }
+ t.push(nums[i]);
+ dfs(i + 1);
+ t.pop();
+ while (i + 1 < n && nums[i] === nums[i + 1]) {
+ i++;
+ }
+ dfs(i + 1);
+ };
+ dfs(0);
+ return ans;
+};
diff --git a/solution/0000-0099/0090.Subsets II/Solution2.cpp b/solution/0000-0099/0090.Subsets II/Solution2.cpp
index 2d5b88450b3e1..397f973163ac5 100644
--- a/solution/0000-0099/0090.Subsets II/Solution2.cpp
+++ b/solution/0000-0099/0090.Subsets II/Solution2.cpp
@@ -1,7 +1,7 @@
class Solution {
public:
vector<vector<int>> subsetsWithDup(vector<int>& nums) {
- sort(nums.begin(), nums.end());
+ ranges::sort(nums);
int n = nums.size();
vector<vector<int>> ans;
for (int mask = 0; mask < 1 << n; ++mask) {
@@ -22,4 +22,4 @@ class Solution {
}
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/0000-0099/0090.Subsets II/Solution2.cs b/solution/0000-0099/0090.Subsets II/Solution2.cs
new file mode 100644
index 0000000000000..52db892cdab0a
--- /dev/null
+++ b/solution/0000-0099/0090.Subsets II/Solution2.cs
@@ -0,0 +1,24 @@
+public class Solution {
+ public IList<IList<int>> SubsetsWithDup(int[] nums) {
+ Array.Sort(nums);
+ int n = nums.Length;
+ IList<IList<int>> ans = new List<IList<int>>();
+ for (int mask = 0; mask < 1 << n; ++mask) {
+ IList<int> t = new List<int>();
+ bool ok = true;
+ for (int i = 0; i < n; ++i) {
+ if ((mask >> i & 1) == 1) {
+ if (i > 0 && (mask >> (i - 1) & 1) == 0 && nums[i] == nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.Add(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.Add(t);
+ }
+ }
+ return ans;
+ }
+}
diff --git a/solution/0000-0099/0090.Subsets II/Solution2.js b/solution/0000-0099/0090.Subsets II/Solution2.js
new file mode 100644
index 0000000000000..c9ff21b919f9e
--- /dev/null
+++ b/solution/0000-0099/0090.Subsets II/Solution2.js
@@ -0,0 +1,26 @@
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+var subsetsWithDup = function (nums) {
+ nums.sort((a, b) => a - b);
+ const n = nums.length;
+ const ans = [];
+ for (let mask = 0; mask < 1 << n; ++mask) {
+ const t = [];
+ let ok = true;
+ for (let i = 0; i < n; ++i) {
+ if (((mask >> i) & 1) === 1) {
+ if (i && ((mask >> (i - 1)) & 1) === 0 && nums[i] === nums[i - 1]) {
+ ok = false;
+ break;
+ }
+ t.push(nums[i]);
+ }
+ }
+ if (ok) {
+ ans.push(t);
+ }
+ }
+ return ans;
+};
From e9533157a93f5d3345a5c265bd6f09ae3e809810 Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Wed, 5 Feb 2025 07:58:28 +0800
Subject: [PATCH 2/3] fix: docs
---
solution/0000-0099/0090.Subsets II/README.md | 16 +++++++--------
.../0000-0099/0090.Subsets II/README_EN.md | 20 +++++++++----------
2 files changed, 18 insertions(+), 18 deletions(-)
diff --git a/solution/0000-0099/0090.Subsets II/README.md b/solution/0000-0099/0090.Subsets II/README.md
index a1e8455409916..5830aabf4caea 100644
--- a/solution/0000-0099/0090.Subsets II/README.md
+++ b/solution/0000-0099/0090.Subsets II/README.md
@@ -59,17 +59,17 @@ tags:
### 方法一：排序 + DFS
-我们可以先对数组 $nums$ 进行排序，方便去重。
+我们可以先对数组 $\textit{nums}$ 进行排序，方便去重。
-然后，我们设计一个函数 $dfs(i)$，表示当前从第 $i$ 个元素开始搜索子集。函数 $dfs(i)$ 的执行逻辑如下：
+然后，我们设计一个函数 $\textit{dfs}(i)$，表示当前从第 $i$ 个元素开始搜索子集。函数 $\textit{dfs}(i)$ 的执行逻辑如下：
如果 $i \geq n$，说明已经搜索完所有元素，将当前子集加入答案数组中，递归结束。
-如果 $i < n$，将第 $i$ 个元素加入子集，执行 $dfs(i + 1)$，然后将第 $i$ 个元素从子集中移除。接下来，我们判断第 $i$ 个元素是否和下一个元素相同，如果相同，则循环跳过该元素，直到找到第一个和第 $i$ 个元素不同的元素，执行 $dfs(i + 1)$。
+如果 $i < n$，将第 $i$ 个元素加入子集，执行 $\textit{dfs}(i + 1)$，然后将第 $i$ 个元素从子集中移除。接下来，我们判断第 $i$ 个元素是否和下一个元素相同，如果相同，则循环跳过该元素，直到找到第一个和第 $i$ 个元素不同的元素，执行 $\textit{dfs}(i + 1)$。
-最后，我们只需要调用 $dfs(0)$，返回答案数组即可。
+最后，我们只需要调用 $\textit{dfs}(0)$，返回答案数组即可。
-时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
+时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
<!-- tabs:start -->
@@ -308,13 +308,13 @@ public class Solution {
### 方法二：排序 + 二进制枚举
-与方法一类似，我们先对数组 $nums$ 进行排序，方便去重。
+与方法一类似，我们先对数组 $\textit{nums}$ 进行排序，方便去重。
-接下来，我们在 $[0, 2^n)$ 的范围内枚举一个二进制数 $mask$，其中 $mask$ 的二进制表示是一个 $n$ 位的位串，如果 $mask$ 的第 $i$ 位为 $1$，表示选择 $nums[i]$，为 $0$ 表示不选择 $nums[i]$。注意，如果 $mask$ 的 $i - 1$ 位为 $0$，且 $nums[i] = nums[i - 1]$，则说明在当前枚举到的方案中，第 $i$ 个元素和第 $i - 1$ 个元素相同，为了避免重复，我们跳过这种情况。否则，我们将 $mask$ 对应的子集加入答案数组中。
+接下来，我们在 $[0, 2^n)$ 的范围内枚举一个二进制数 $\textit{mask}$，其中 $\textit{mask}$ 的二进制表示是一个 $n$ 位的位串，如果 $\textit{mask}$ 的第 $i$ 位为 $1$，表示选择 $\textit{nums}[i]$，为 $0$ 表示不选择 $\textit{nums}[i]$。注意，如果 $\textit{mask}$ 的 $i - 1$ 位为 $0$，且 $\textit{nums}[i] = \textit{nums}[i - 1]$，则说明在当前枚举到的方案中，第 $i$ 个元素和第 $i - 1$ 个元素相同，为了避免重复，我们跳过这种情况。否则，我们将 $\textit{mask}$ 对应的子集加入答案数组中。
枚举结束后，我们返回答案数组即可。
-时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(n)$。其中 $n$ 是数组的长度。
+时间复杂度 $O(n \times 2^n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{nums}$ 的长度。
<!-- tabs:start -->
diff --git a/solution/0000-0099/0090.Subsets II/README_EN.md b/solution/0000-0099/0090.Subsets II/README_EN.md
index cf0c5634e19a1..f23aca40907b0 100644
--- a/solution/0000-0099/0090.Subsets II/README_EN.md
+++ b/solution/0000-0099/0090.Subsets II/README_EN.md
@@ -46,17 +46,17 @@ tags:
### Solution 1: Sorting + DFS
-We can first sort the array $nums$ to facilitate deduplication.
+We can first sort the array $\textit{nums}$ to facilitate deduplication.
-Then, we design a function $dfs(i)$, which represents searching for subsets starting from the $i$-th element. The execution logic of the function $dfs(i)$ is as follows:
+Then, we design a function $\textit{dfs}(i)$, which represents the current search for subsets starting from the $i$-th element. The execution logic of the function $\textit{dfs}(i)$ is as follows:
-If $i \geq n$, it means that all elements have been searched, and the current subset is added to the answer array, and the recursion ends.
+If $i \geq n$, it means all elements have been searched, add the current subset to the answer array, and end the recursion.
-If $i < n$, add the $i$-th element to the subset, execute $dfs(i + 1)$, and then remove the $i$-th element from the subset. Next, we judge whether the $i$-th element is the same as the next element. If it is the same, we loop to skip this element until we find the first element that is different from the $i$-th element, and execute $dfs(i + 1)$.
+If $i < n$, add the $i$-th element to the subset, execute $\textit{dfs}(i + 1)$, then remove the $i$-th element from the subset. Next, we check if the $i$-th element is the same as the next element. If they are the same, skip the element in a loop until we find the first element different from the $i$-th element, then execute $\textit{dfs}(i + 1)$.
-Finally, we only need to call $dfs(0)$ and return the answer array.
+Finally, we only need to call $\textit{dfs}(0)$ and return the answer array.
-The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
+The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
<!-- tabs:start -->
@@ -295,13 +295,13 @@ public class Solution {
### Solution 2: Sorting + Binary Enumeration
-Similar to Solution 1, we first sort the array $nums$ to facilitate deduplication.
+Similar to Solution 1, we first sort the array $\textit{nums}$ to facilitate deduplication.
-Next, we enumerate a binary number $mask$ in the range of $[0, 2^n)$, where the binary representation of $mask$ is an $n$-bit bit string. If the $i$-th bit of $mask$ is $1$, it means to select $nums[i]$, and $0$ means not to select $nums[i]$. Note that if the $i - 1$ bit of $mask$ is $0$, and $nums[i] = nums[i - 1]$, it means that in the current enumerated scheme, the $i$-th element and the $i - 1$-th element are the same. To avoid repetition, we skip this situation. Otherwise, we add the subset corresponding to $mask$ to the answer array.
+Next, we enumerate a binary number $\textit{mask}$ in the range $[0, 2^n)$, where the binary representation of $\textit{mask}$ is an $n$-bit bit string. If the $i$-th bit of $\textit{mask}$ is $1$, it means selecting $\textit{nums}[i]$, and $0$ means not selecting $\textit{nums}[i]$. Note that if the $(i - 1)$-th bit of $\textit{mask}$ is $0$ and $\textit{nums}[i] = \textit{nums}[i - 1]$, it means that the $i$-th element is the same as the $(i - 1)$-th element in the current enumeration scheme. To avoid duplication, we skip this case. Otherwise, we add the subset corresponding to $\textit{mask}$ to the answer array.
-After the enumeration ends, we return the answer array.
+After the enumeration, we return the answer array.
-The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array.
+The time complexity is $O(n \times 2^n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
<!-- tabs:start -->
From 801a7e8b1944ee188b215b35123865adee1c8c86 Mon Sep 17 00:00:00 2001
From: yanglbme <szuyanglb@outlook.com>
Date: Wed, 5 Feb 2025 08:06:32 +0800
Subject: [PATCH 3/3] fix: update solutions
---
.../0009.Palindrome Number/README.md | 34 +++++++++----------
.../0009.Palindrome Number/README_EN.md | 34 +++++++++----------
.../0009.Palindrome Number/Solution.cs | 12 +++----
.../0009.Palindrome Number/Solution.php | 17 ++++++----
.../0009.Palindrome Number/Solution.rs | 21 ++++--------
5 files changed, 56 insertions(+), 62 deletions(-)
diff --git a/solution/0000-0099/0009.Palindrome Number/README.md b/solution/0000-0099/0009.Palindrome Number/README.md
index ee1c738dee14d..5293dccce090b 100644
--- a/solution/0000-0099/0009.Palindrome Number/README.md
+++ b/solution/0000-0099/0009.Palindrome Number/README.md
@@ -177,13 +177,12 @@ function isPalindrome(x: number): boolean {
```rust
impl Solution {
pub fn is_palindrome(mut x: i32) -> bool {
- if x < 0 || (x % 10 == 0 && x != 0) {
+ if x < 0 || (x != 0 && x % 10 == 0) {
return false;
}
let mut y = 0;
while x > y {
- y *= 10;
- y += x % 10;
+ y = y * 10 + x % 10;
x /= 10;
}
x == y || x == y / 10
@@ -213,17 +212,13 @@ var isPalindrome = function (x) {
#### C#
```cs
-public class Solution
-{
- public bool IsPalindrome(int x)
- {
- if (x < 0 || (x > 0 && x % 10 == 0))
- {
+public class Solution {
+ public bool IsPalindrome(int x) {
+ if (x < 0 || (x > 0 && x % 10 == 0)) {
return false;
}
int y = 0;
- for (; y < x; x /= 10)
- {
+ for (; y < x; x /= 10) {
y = y * 10 + x % 10;
}
return x == y || x == y / 10;
@@ -236,14 +231,19 @@ public class Solution
```php
class Solution {
/**
- * @param int $x
- * @return boolean
+ * @param Integer $x
+ * @return Boolean
*/
-
function isPalindrome($x) {
- $str = (string) $x;
- $str_reverse = strrev($str);
- return $str === $str_reverse;
+ if ($x < 0 || ($x && $x % 10 == 0)) {
+ return false;
+ }
+ $y = 0;
+ while ($x > $y) {
+ $y = $y * 10 + ($x % 10);
+ $x = (int) ($x / 10);
+ }
+ return $x == $y || $x == (int) ($y / 10);
}
}
```
diff --git a/solution/0000-0099/0009.Palindrome Number/README_EN.md b/solution/0000-0099/0009.Palindrome Number/README_EN.md
index 5616eee00755d..8b0ac2e383ca0 100644
--- a/solution/0000-0099/0009.Palindrome Number/README_EN.md
+++ b/solution/0000-0099/0009.Palindrome Number/README_EN.md
@@ -169,13 +169,12 @@ function isPalindrome(x: number): boolean {
```rust
impl Solution {
pub fn is_palindrome(mut x: i32) -> bool {
- if x < 0 || (x % 10 == 0 && x != 0) {
+ if x < 0 || (x != 0 && x % 10 == 0) {
return false;
}
let mut y = 0;
while x > y {
- y *= 10;
- y += x % 10;
+ y = y * 10 + x % 10;
x /= 10;
}
x == y || x == y / 10
@@ -205,17 +204,13 @@ var isPalindrome = function (x) {
#### C#
```cs
-public class Solution
-{
- public bool IsPalindrome(int x)
- {
- if (x < 0 || (x > 0 && x % 10 == 0))
- {
+public class Solution {
+ public bool IsPalindrome(int x) {
+ if (x < 0 || (x > 0 && x % 10 == 0)) {
return false;
}
int y = 0;
- for (; y < x; x /= 10)
- {
+ for (; y < x; x /= 10) {
y = y * 10 + x % 10;
}
return x == y || x == y / 10;
@@ -228,14 +223,19 @@ public class Solution
```php
class Solution {
/**
- * @param int $x
- * @return boolean
+ * @param Integer $x
+ * @return Boolean
*/
-
function isPalindrome($x) {
- $str = (string) $x;
- $str_reverse = strrev($str);
- return $str === $str_reverse;
+ if ($x < 0 || ($x && $x % 10 == 0)) {
+ return false;
+ }
+ $y = 0;
+ while ($x > $y) {
+ $y = $y * 10 + ($x % 10);
+ $x = (int) ($x / 10);
+ }
+ return $x == $y || $x == (int) ($y / 10);
}
}
```
diff --git a/solution/0000-0099/0009.Palindrome Number/Solution.cs b/solution/0000-0099/0009.Palindrome Number/Solution.cs
index b58946514c161..f530d9156371b 100644
--- a/solution/0000-0099/0009.Palindrome Number/Solution.cs
+++ b/solution/0000-0099/0009.Palindrome Number/Solution.cs
@@ -1,14 +1,10 @@
-public class Solution
-{
- public bool IsPalindrome(int x)
- {
- if (x < 0 || (x > 0 && x % 10 == 0))
- {
+public class Solution {
+ public bool IsPalindrome(int x) {
+ if (x < 0 || (x > 0 && x % 10 == 0)) {
return false;
}
int y = 0;
- for (; y < x; x /= 10)
- {
+ for (; y < x; x /= 10) {
y = y * 10 + x % 10;
}
return x == y || x == y / 10;
diff --git a/solution/0000-0099/0009.Palindrome Number/Solution.php b/solution/0000-0099/0009.Palindrome Number/Solution.php
index 9fd8c6654ec95..c4042d56fbed1 100644
--- a/solution/0000-0099/0009.Palindrome Number/Solution.php
+++ b/solution/0000-0099/0009.Palindrome Number/Solution.php
@@ -1,12 +1,17 @@
class Solution {
/**
- * @param int $x
- * @return boolean
+ * @param Integer $x
+ * @return Boolean
*/
-
function isPalindrome($x) {
- $str = (string) $x;
- $str_reverse = strrev($str);
- return $str === $str_reverse;
+ if ($x < 0 || ($x && $x % 10 == 0)) {
+ return false;
+ }
+ $y = 0;
+ while ($x > $y) {
+ $y = $y * 10 + ($x % 10);
+ $x = (int) ($x / 10);
+ }
+ return $x == $y || $x == (int) ($y / 10);
}
}
diff --git a/solution/0000-0099/0009.Palindrome Number/Solution.rs b/solution/0000-0099/0009.Palindrome Number/Solution.rs
index 7163690151d08..9275d20a011d7 100644
--- a/solution/0000-0099/0009.Palindrome Number/Solution.rs
+++ b/solution/0000-0099/0009.Palindrome Number/Solution.rs
@@ -1,20 +1,13 @@
impl Solution {
- pub fn is_palindrome(x: i32) -> bool {
- if x < 0 {
+ pub fn is_palindrome(mut x: i32) -> bool {
+ if x < 0 || (x != 0 && x % 10 == 0) {
return false;
}
- let s = x.to_string();
- let bs = s.as_bytes();
- let n = bs.len();
- let mut l = 0;
- let mut r = n - 1;
- while l < r {
- if bs[l] != bs[r] {
- return false;
- }
- l += 1;
- r -= 1;
+ let mut y = 0;
+ while x > y {
+ y = y * 10 + x % 10;
+ x /= 10;
}
- true
+ x == y || x == y / 10
}
}