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Commit f31828a

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feat: add solutions to lc problems: No.2485~2487 (doocs#2175)
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‎solution/2400-2499/2485.Find the Pivot Integer/README_EN.md‎

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@@ -46,6 +46,34 @@
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## Solutions
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49+
**Solution 1: Enumeration**
50+
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We can directly enumerate $x$ in the range of $[1,..n],ドル and check whether the following equation holds. If it holds, then $x$ is the pivot integer, and we can directly return $x$.
52+
53+
$$
54+
(1 + x) \times x = (x + n) \times (n - x + 1)
55+
$$
56+
57+
The time complexity is $O(n),ドル where $n$ is the given positive integer $n$. The space complexity is $O(1)$.
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**Solution 2: Mathematics**
60+
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We can transform the above equation to get:
62+
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$$
64+
n \times (n + 1) = 2 \times x^2
65+
$$
66+
67+
That is:
68+
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$$
70+
x = \sqrt{\frac{n \times (n + 1)}{2}}
71+
$$
72+
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If $x$ is an integer, then $x$ is the pivot integer, otherwise there is no pivot integer.
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The time complexity is $O(1),ドル and the space complexity is $O(1)$.
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4977
<!-- tabs:start -->
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### **Python3**

‎solution/2400-2499/2486.Append Characters to String to Make Subsequence/README.md‎

Lines changed: 23 additions & 6 deletions
Original file line numberDiff line numberDiff line change
@@ -57,7 +57,7 @@
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**方法一:双指针**
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定义两个指针 $i$ 和 $j,ドル分别指向字符串 $s$ 和 $t$ 的首字符。遍历字符串 $t,ドル当 $s[i] \neq t[j]$ 时,指针 $i$ 后移,直到 $s[i] = t[j]$ 或者 $i$ 到达字符串 $s$ 的末尾。如果 $i$ 到达字符串 $s$ 的末尾,说明 $t$ 中的字符 $t[j]$ 无法在 $s$ 中找到对应的字符,返回 $t$ 中剩余的字符数。否则,将指针 $i$ 和 $j$ 同时后移,继续遍历字符串 $t$。
60+
我们定义两个指针 $i$ 和 $j,ドル分别指向字符串 $s$ 和 $t$ 的首字符。遍历字符串 $t,ドル当 $s[i] \neq t[j]$ 时,指针 $i$ 后移,直到 $s[i] = t[j]$ 或者 $i$ 到达字符串 $s$ 的末尾。如果 $i$ 到达字符串 $s$ 的末尾,说明 $t$ 中的字符 $t[j]$ 无法在 $s$ 中找到对应的字符,返回 $t$ 中剩余的字符数。否则,将指针 $i$ 和 $j$ 同时后移,继续遍历字符串 $t$。
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时间复杂度 $(m + n),ドル空间复杂度 $O(1)$。其中 $m$ 和 $n$ 分别是字符串 $s$ 和 $t$ 的长度。
6363

@@ -70,13 +70,12 @@
7070
```python
7171
class Solution:
7272
def appendCharacters(self, s: str, t: str) -> int:
73-
m, n = len(s), len(t)
74-
i = 0
75-
for j in range(n):
76-
while i < m and s[i] != t[j]:
73+
i, m = 0, len(s)
74+
for j, c in enumerate(t):
75+
while i < m and s[i] != c:
7776
i += 1
7877
if i == m:
79-
return n - j
78+
return len(t) - j
8079
i += 1
8180
return 0
8281
```
@@ -139,6 +138,24 @@ func appendCharacters(s string, t string) int {
139138
}
140139
```
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### **TypeScript**
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```ts
144+
function appendCharacters(s: string, t: string): number {
145+
const [m, n] = [s.length, t.length];
146+
for (let i = 0, j = 0; j < n; ++j) {
147+
while (i < m && s[i] !== t[j]) {
148+
++i;
149+
}
150+
if (i === m) {
151+
return n - j;
152+
}
153+
++i;
154+
}
155+
return 0;
156+
}
157+
```
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142159
### **...**
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```

‎solution/2400-2499/2486.Append Characters to String to Make Subsequence/README_EN.md‎

Lines changed: 28 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -49,20 +49,25 @@ It can be shown that appending any 4 characters to the end of s will never make
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## Solutions
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**Solution 1: Two Pointers**
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We define two pointers $i$ and $j,ドル pointing to the first characters of strings $s$ and $t$ respectively. We traverse string $t,ドル when $s[i] \neq t[j],ドル we move pointer $i$ forward until $s[i] = t[j]$ or $i$ reaches the end of string $s$. If $i$ reaches the end of string $s,ドル it means that the character $t[j]$ in $t$ cannot find the corresponding character in $s,ドル so we return the remaining number of characters in $t$. Otherwise, we move both pointers $i$ and $j$ forward and continue to traverse string $t$.
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The time complexity is $O(m + n),ドル and the space complexity is $O(1)$. Where $m$ and $n$ are the lengths of strings $s$ and $t$ respectively.
57+
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<!-- tabs:start -->
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### **Python3**
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```python
5763
class Solution:
5864
def appendCharacters(self, s: str, t: str) -> int:
59-
m, n = len(s), len(t)
60-
i = 0
61-
for j in range(n):
62-
while i < m and s[i] != t[j]:
65+
i, m = 0, len(s)
66+
for j, c in enumerate(t):
67+
while i < m and s[i] != c:
6368
i += 1
6469
if i == m:
65-
return n - j
70+
return len(t) - j
6671
i += 1
6772
return 0
6873
```
@@ -123,6 +128,24 @@ func appendCharacters(s string, t string) int {
123128
}
124129
```
125130

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### **TypeScript**
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133+
```ts
134+
function appendCharacters(s: string, t: string): number {
135+
const [m, n] = [s.length, t.length];
136+
for (let i = 0, j = 0; j < n; ++j) {
137+
while (i < m && s[i] !== t[j]) {
138+
++i;
139+
}
140+
if (i === m) {
141+
return n - j;
142+
}
143+
++i;
144+
}
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return 0;
146+
}
147+
```
148+
126149
### **...**
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128151
```
Lines changed: 10 additions & 11 deletions
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,10 @@
1-
class Solution:
2-
def appendCharacters(self, s: str, t: str) -> int:
3-
m, n = len(s), len(t)
4-
i = 0
5-
for j in range(n):
6-
while i < m and s[i] != t[j]:
7-
i += 1
8-
if i == m:
9-
return n - j
10-
i += 1
11-
return 0
1+
class Solution:
2+
def appendCharacters(self, s: str, t: str) -> int:
3+
i, m = 0, len(s)
4+
for j, c in enumerate(t):
5+
while i < m and s[i] != c:
6+
i += 1
7+
if i == m:
8+
return len(t) - j
9+
i += 1
10+
return 0
Lines changed: 13 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,13 @@
1+
function appendCharacters(s: string, t: string): number {
2+
const [m, n] = [s.length, t.length];
3+
for (let i = 0, j = 0; j < n; ++j) {
4+
while (i < m && s[i] !== t[j]) {
5+
++i;
6+
}
7+
if (i === m) {
8+
return n - j;
9+
}
10+
++i;
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}
12+
return 0;
13+
}

‎solution/2400-2499/2487.Remove Nodes From Linked List/README.md‎

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@@ -56,6 +56,12 @@
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是链表的长度。
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我们也可以不使用数组 $nums,ドル直接遍历链表,维护一个从栈底到栈顶单调递减的栈 $stk,ドル如果当前元素比栈顶元素大,则将栈顶元素出栈,直到当前元素小于等于栈顶元素。然后,如果栈不为空,则将栈顶元素的 $next$ 指针指向当前元素,否则将答案链表的虚拟头节点的 $next$ 指针指向当前元素。最后,将当前元素入栈,继续遍历链表。
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遍历结束后,将虚拟头节点的 $next$ 指针作为答案返回。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是链表的长度。
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<!-- tabs:start -->
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### **Python3**
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8793
return dummy.next
8894
```
8995

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```python
97+
# Definition for singly-linked list.
98+
# class ListNode:
99+
# def __init__(self, val=0, next=None):
100+
# self.val = val
101+
# self.next = next
102+
class Solution:
103+
def removeNodes(self, head: Optional[ListNode]) -> Optional[ListNode]:
104+
dummy = ListNode(next=head)
105+
cur = head
106+
stk = []
107+
while cur:
108+
while stk and stk[-1].val < cur.val:
109+
stk.pop()
110+
if stk:
111+
stk[-1].next = cur
112+
else:
113+
dummy.next = cur
114+
stk.append(cur)
115+
cur = cur.next
116+
return dummy.next
117+
```
118+
90119
### **Java**
91120

92121
<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -127,6 +156,37 @@ class Solution {
127156
}
128157
```
129158

159+
```java
160+
/**
161+
* Definition for singly-linked list.
162+
* public class ListNode {
163+
* int val;
164+
* ListNode next;
165+
* ListNode() {}
166+
* ListNode(int val) { this.val = val; }
167+
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
168+
* }
169+
*/
170+
class Solution {
171+
public ListNode removeNodes(ListNode head) {
172+
ListNode dummy = new ListNode(0, head);
173+
Deque<ListNode> stk = new ArrayDeque<>();
174+
for (ListNode cur = head; cur != null; cur = cur.next) {
175+
while (!stk.isEmpty() && stk.peekLast().val < cur.val) {
176+
stk.pollLast();
177+
}
178+
if (!stk.isEmpty()) {
179+
stk.peekLast().next = cur;
180+
} else {
181+
dummy.next = cur;
182+
}
183+
stk.offerLast(cur);
184+
}
185+
return dummy.next;
186+
}
187+
}
188+
```
189+
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### **C++**
131191

132192
```cpp
@@ -166,6 +226,39 @@ public:
166226
};
167227
```
168228
229+
```cpp
230+
/**
231+
* Definition for singly-linked list.
232+
* struct ListNode {
233+
* int val;
234+
* ListNode *next;
235+
* ListNode() : val(0), next(nullptr) {}
236+
* ListNode(int x) : val(x), next(nullptr) {}
237+
* ListNode(int x, ListNode *next) : val(x), next(next) {}
238+
* };
239+
*/
240+
class Solution {
241+
public:
242+
ListNode* removeNodes(ListNode* head) {
243+
ListNode* dummy = new ListNode(0, head);
244+
ListNode* cur = head;
245+
vector<ListNode*> stk;
246+
for (ListNode* cur = head; cur; cur = cur->next) {
247+
while (stk.size() && stk.back()->val < cur->val) {
248+
stk.pop_back();
249+
}
250+
if (stk.size()) {
251+
stk.back()->next = cur;
252+
} else {
253+
dummy->next = cur;
254+
}
255+
stk.push_back(cur);
256+
}
257+
return dummy->next;
258+
}
259+
};
260+
```
261+
169262
### **Go**
170263

171264
```go
@@ -199,6 +292,32 @@ func removeNodes(head *ListNode) *ListNode {
199292
}
200293
```
201294

295+
```go
296+
/**
297+
* Definition for singly-linked list.
298+
* type ListNode struct {
299+
* Val int
300+
* Next *ListNode
301+
* }
302+
*/
303+
func removeNodes(head *ListNode) *ListNode {
304+
dummy := &ListNode{Next: head}
305+
stk := []*ListNode{}
306+
for cur := head; cur != nil; cur = cur.Next {
307+
for len(stk) > 0 && stk[len(stk)-1].Val < cur.Val {
308+
stk = stk[:len(stk)-1]
309+
}
310+
if len(stk) > 0 {
311+
stk[len(stk)-1].Next = cur
312+
} else {
313+
dummy.Next = cur
314+
}
315+
stk = append(stk, cur)
316+
}
317+
return dummy.Next
318+
}
319+
```
320+
202321
### **TypeScript**
203322

204323
```ts
@@ -236,6 +355,37 @@ function removeNodes(head: ListNode | null): ListNode | null {
236355
}
237356
```
238357

358+
```ts
359+
/**
360+
* Definition for singly-linked list.
361+
* class ListNode {
362+
* val: number
363+
* next: ListNode | null
364+
* constructor(val?: number, next?: ListNode | null) {
365+
* this.val = (val===undefined ? 0 : val)
366+
* this.next = (next===undefined ? null : next)
367+
* }
368+
* }
369+
*/
370+
371+
function removeNodes(head: ListNode | null): ListNode | null {
372+
const dummy = new ListNode(0, head);
373+
const stk: ListNode[] = [];
374+
for (let cur = head; cur; cur = cur.next) {
375+
while (stk.length && stk.at(-1)!.val < cur.val) {
376+
stk.pop();
377+
}
378+
if (stk.length) {
379+
stk.at(-1)!.next = cur;
380+
} else {
381+
dummy.next = cur;
382+
}
383+
stk.push(cur);
384+
}
385+
return dummy.next;
386+
}
387+
```
388+
239389
### **...**
240390

241391
```

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