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Commit 08f9dc0

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feat: add solutions to lcci problem: No.02.02 (doocs#2306)
No.02.02.Kth Node From End of List
1 parent 3aff15b commit 08f9dc0

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7 files changed

+130
-26
lines changed

7 files changed

+130
-26
lines changed

‎lcci/02.02.Kth Node From End of List/README.md‎

Lines changed: 46 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -20,7 +20,11 @@
2020

2121
## 解法
2222

23-
### 方法一
23+
### 方法一:快慢指针
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25+
我们定义两个指针 `slow` 和 `fast`,初始时都指向链表头节点 `head`。然后 `fast` 指针先向前移动 $k$ 步,然后 `slow` 和 `fast` 指针同时向前移动,直到 `fast` 指针指向链表末尾。此时 `slow` 指针指向的节点就是倒数第 $k$ 个节点。
26+
27+
时间复杂度 $O(n),ドル其中 $n$ 是链表的长度。空间复杂度 $O(1)$。
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2529
<!-- tabs:start -->
2630

@@ -38,7 +42,8 @@ class Solution:
3842
for _ in range(k):
3943
fast = fast.next
4044
while fast:
41-
slow, fast = slow.next, fast.next
45+
slow = slow.next
46+
fast = fast.next
4247
return slow.val
4348
```
4449

@@ -80,7 +85,7 @@ public:
8085
int kthToLast(ListNode* head, int k) {
8186
ListNode* fast = head;
8287
ListNode* slow = head;
83-
while (k-- > 0) {
88+
while (k--) {
8489
fast = fast->next;
8590
}
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while (fast) {
@@ -93,9 +98,16 @@ public:
9398
```
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```go
101+
/**
102+
* Definition for singly-linked list.
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* type ListNode struct {
104+
* Val int
105+
* Next *ListNode
106+
* }
107+
*/
96108
func kthToLast(head *ListNode, k int) int {
97109
slow, fast := head, head
98-
for i := 0; i < k; i++ {
110+
for ; k > 0; k-- {
99111
fast = fast.Next
100112
}
101113
for fast != nil {
@@ -106,6 +118,32 @@ func kthToLast(head *ListNode, k int) int {
106118
}
107119
```
108120

121+
```ts
122+
/**
123+
* Definition for singly-linked list.
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* class ListNode {
125+
* val: number
126+
* next: ListNode | null
127+
* constructor(val?: number, next?: ListNode | null) {
128+
* this.val = (val===undefined ? 0 : val)
129+
* this.next = (next===undefined ? null : next)
130+
* }
131+
* }
132+
*/
133+
134+
function kthToLast(head: ListNode | null, k: number): number {
135+
let [slow, fast] = [head, head];
136+
while (k--) {
137+
fast = fast.next;
138+
}
139+
while (fast !== null) {
140+
slow = slow.next;
141+
fast = fast.next;
142+
}
143+
return slow.val;
144+
}
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```
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109147
```rust
110148
// Definition for singly-linked list.
111149
// #[derive(PartialEq, Eq, Clone, Debug)]
@@ -153,14 +191,13 @@ impl Solution {
153191
* @return {number}
154192
*/
155193
var kthToLast = function (head, k) {
156-
let fast = head,
157-
slow = head;
158-
for (let i = 0; i < k; i++) {
194+
let [slow, fast] = [head, head];
195+
while (k--) {
159196
fast = fast.next;
160197
}
161-
while (fast != null) {
162-
fast = fast.next;
198+
while (fast !== null) {
163199
slow = slow.next;
200+
fast = fast.next;
164201
}
165202
return slow.val;
166203
};

‎lcci/02.02.Kth Node From End of List/README_EN.md‎

Lines changed: 46 additions & 9 deletions
Original file line numberDiff line numberDiff line change
@@ -22,7 +22,11 @@
2222

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## Solutions
2424

25-
### Solution 1
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### Solution 1: Two Pointers
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We define two pointers `slow` and `fast`, both initially pointing to the head node `head`. Then the `fast` pointer moves forward $k$ steps first, and then the `slow` and `fast` pointers move forward together until the `fast` pointer points to the end of the list. At this point, the node pointed to by the `slow` pointer is the $k$-th node from the end of the list.
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The time complexity is $O(n),ドル where $n$ is the length of the list. The space complexity is $O(1)$.
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<!-- tabs:start -->
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@@ -40,7 +44,8 @@ class Solution:
4044
for _ in range(k):
4145
fast = fast.next
4246
while fast:
43-
slow, fast = slow.next, fast.next
47+
slow = slow.next
48+
fast = fast.next
4449
return slow.val
4550
```
4651

@@ -82,7 +87,7 @@ public:
8287
int kthToLast(ListNode* head, int k) {
8388
ListNode* fast = head;
8489
ListNode* slow = head;
85-
while (k-- > 0) {
90+
while (k--) {
8691
fast = fast->next;
8792
}
8893
while (fast) {
@@ -95,9 +100,16 @@ public:
95100
```
96101
97102
```go
103+
/**
104+
* Definition for singly-linked list.
105+
* type ListNode struct {
106+
* Val int
107+
* Next *ListNode
108+
* }
109+
*/
98110
func kthToLast(head *ListNode, k int) int {
99111
slow, fast := head, head
100-
for i := 0; i < k; i++ {
112+
for ; k > 0; k-- {
101113
fast = fast.Next
102114
}
103115
for fast != nil {
@@ -108,6 +120,32 @@ func kthToLast(head *ListNode, k int) int {
108120
}
109121
```
110122

123+
```ts
124+
/**
125+
* Definition for singly-linked list.
126+
* class ListNode {
127+
* val: number
128+
* next: ListNode | null
129+
* constructor(val?: number, next?: ListNode | null) {
130+
* this.val = (val===undefined ? 0 : val)
131+
* this.next = (next===undefined ? null : next)
132+
* }
133+
* }
134+
*/
135+
136+
function kthToLast(head: ListNode | null, k: number): number {
137+
let [slow, fast] = [head, head];
138+
while (k--) {
139+
fast = fast.next;
140+
}
141+
while (fast !== null) {
142+
slow = slow.next;
143+
fast = fast.next;
144+
}
145+
return slow.val;
146+
}
147+
```
148+
111149
```rust
112150
// Definition for singly-linked list.
113151
// #[derive(PartialEq, Eq, Clone, Debug)]
@@ -155,14 +193,13 @@ impl Solution {
155193
* @return {number}
156194
*/
157195
var kthToLast = function (head, k) {
158-
let fast = head,
159-
slow = head;
160-
for (let i = 0; i < k; i++) {
196+
let [slow, fast] = [head, head];
197+
while (k--) {
161198
fast = fast.next;
162199
}
163-
while (fast != null) {
164-
fast = fast.next;
200+
while (fast !== null) {
165201
slow = slow.next;
202+
fast = fast.next;
166203
}
167204
return slow.val;
168205
};

‎lcci/02.02.Kth Node From End of List/Solution.cpp‎

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -11,7 +11,7 @@ class Solution {
1111
int kthToLast(ListNode* head, int k) {
1212
ListNode* fast = head;
1313
ListNode* slow = head;
14-
while (k-- > 0) {
14+
while (k--) {
1515
fast = fast->next;
1616
}
1717
while (fast) {

‎lcci/02.02.Kth Node From End of List/Solution.go‎

Lines changed: 8 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -1,6 +1,13 @@
1+
/**
2+
* Definition for singly-linked list.
3+
* type ListNode struct {
4+
* Val int
5+
* Next *ListNode
6+
* }
7+
*/
18
func kthToLast(head *ListNode, k int) int {
29
slow, fast := head, head
3-
for i:=0; i<k; i++ {
10+
for ; k>0; k-- {
411
fast = fast.Next
512
}
613
for fast != nil {

‎lcci/02.02.Kth Node From End of List/Solution.js‎

Lines changed: 4 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -11,14 +11,13 @@
1111
* @return {number}
1212
*/
1313
var kthToLast = function (head, k) {
14-
let fast = head,
15-
slow = head;
16-
for (let i = 0; i < k; i++) {
14+
let [slow, fast] = [head, head];
15+
while (k--) {
1716
fast = fast.next;
1817
}
19-
while (fast != null) {
20-
fast = fast.next;
18+
while (fast !== null) {
2119
slow = slow.next;
20+
fast = fast.next;
2221
}
2322
return slow.val;
2423
};

‎lcci/02.02.Kth Node From End of List/Solution.py‎

Lines changed: 2 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -11,5 +11,6 @@ def kthToLast(self, head: ListNode, k: int) -> int:
1111
for _ in range(k):
1212
fast = fast.next
1313
while fast:
14-
slow, fast = slow.next, fast.next
14+
slow = slow.next
15+
fast = fast.next
1516
return slow.val
Lines changed: 23 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,23 @@
1+
/**
2+
* Definition for singly-linked list.
3+
* class ListNode {
4+
* val: number
5+
* next: ListNode | null
6+
* constructor(val?: number, next?: ListNode | null) {
7+
* this.val = (val===undefined ? 0 : val)
8+
* this.next = (next===undefined ? null : next)
9+
* }
10+
* }
11+
*/
12+
13+
function kthToLast(head: ListNode | null, k: number): number {
14+
let [slow, fast] = [head, head];
15+
while (k--) {
16+
fast = fast.next;
17+
}
18+
while (fast !== null) {
19+
slow = slow.next;
20+
fast = fast.next;
21+
}
22+
return slow.val;
23+
}

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