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feat: update leetcode solutions: 0725. Split Linked List in Parts
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‎solution/0700-0799/0725.Split Linked List in Parts/README.md

+68-2
Original file line numberDiff line numberDiff line change
@@ -54,22 +54,88 @@ root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3
5454

5555
<!-- 这里可写通用的实现逻辑 -->
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57+
先遍历链表，统计结点总个数。
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59+
接着将链表拆分，`width` 表示每一部分至少含有的结点个数，而 `remainder` 表示前 `remainder` 部分，每一部分多出一个数。
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然后遍历链表，依次拆出每一部分，添加到结果数组 `res` 即可。
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5763
<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
64-
70+
# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
74+
# self.next = None
75+
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class Solution:
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def splitListToParts(self, root: ListNode, k: int) -> List[ListNode]:
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n, cur = 0, root
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while cur:
80+
n += 1
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cur = cur.next
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cur = root
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width, remainder = divmod(n, k)
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res = [None for _ in range(k)]
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for i in range(k):
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head = cur
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for j in range(width + (i < remainder) - 1):
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if cur:
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cur = cur.next
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if cur:
91+
cur.next, cur = None, cur.next
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res[i] = head
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return res
6594
```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
72-
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/**
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* Definition for singly-linked list.
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* public class ListNode {
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* int val;
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* ListNode next;
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* ListNode(int x) { val = x; }
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* }
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*/
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class Solution {
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public ListNode[] splitListToParts(ListNode root, int k) {
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int n = 0;
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ListNode cur = root;
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while (cur != null) {
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++n;
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cur = cur.next;
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}
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// width 表示每一部分至少含有的结点个数
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// remainder 表示前 remainder 部分，每一部分多出一个数
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int width = n / k, remainder = n % k;
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ListNode[] res = new ListNode[k];
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cur = root;
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for (int i = 0; i < k; ++i) {
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ListNode head = cur;
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for (int j = 0; j < width + ((i < remainder) ? 1 : 0) - 1; ++j) {
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if (cur != null) {
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cur = cur.next;
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}
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}
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if (cur != null) {
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ListNode t = cur.next;
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cur.next = null;
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cur = t;
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}
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res[i] = head;
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}
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return res;
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}
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}
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```
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### **...**

‎solution/0700-0799/0725.Split Linked List in Parts/README_EN.md

+60-2
Original file line numberDiff line numberDiff line change
@@ -93,13 +93,71 @@ The input has been split into consecutive parts with size difference at most 1,
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### **Python3**
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```python
96-
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
99+
# self.val = x
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# self.next = None
101+
102+
class Solution:
103+
def splitListToParts(self, root: ListNode, k: int) -> List[ListNode]:
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n, cur = 0, root
105+
while cur:
106+
n += 1
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cur = cur.next
108+
cur = root
109+
width, remainder = divmod(n, k)
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res = [None for _ in range(k)]
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for i in range(k):
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head = cur
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for j in range(width + (i < remainder) - 1):
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if cur:
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cur = cur.next
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if cur:
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cur.next, cur = None, cur.next
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res[i] = head
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return res
97120
```
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### **Java**
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```java
102-
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/**
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* Definition for singly-linked list.
127+
* public class ListNode {
128+
* int val;
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* ListNode next;
130+
* ListNode(int x) { val = x; }
131+
* }
132+
*/
133+
class Solution {
134+
public ListNode[] splitListToParts(ListNode root, int k) {
135+
int n = 0;
136+
ListNode cur = root;
137+
while (cur != null) {
138+
++n;
139+
cur = cur.next;
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}
141+
int width = n / k, remainder = n % k;
142+
ListNode[] res = new ListNode[k];
143+
cur = root;
144+
for (int i = 0; i < k; ++i) {
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ListNode head = cur;
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for (int j = 0; j < width + ((i < remainder) ? 1 : 0) - 1; ++j) {
147+
if (cur != null) {
148+
cur = cur.next;
149+
}
150+
}
151+
if (cur != null) {
152+
ListNode t = cur.next;
153+
cur.next = null;
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cur = t;
155+
}
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res[i] = head;
157+
}
158+
return res;
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}
160+
}
103161
```
104162

105163
### **...**

‎solution/0700-0799/0725.Split Linked List in Parts/Solution.java

+30-23
Original file line numberDiff line numberDiff line change
@@ -1,31 +1,38 @@
1+
/**
2+
* Definition for singly-linked list.
3+
* public class ListNode {
4+
* int val;
5+
* ListNode next;
6+
* ListNode(int x) { val = x; }
7+
* }
8+
*/
19
class Solution {
210
public ListNode[] splitListToParts(ListNode root, int k) {
11+
int n = 0;
12+
ListNode cur = root;
13+
while (cur != null) {
14+
++n;
15+
cur = cur.next;
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}
17+
// width 表示每一部分至少含有的结点个数
18+
// remainder 表示前 remainder 部分，每一部分多出一个数
19+
int width = n / k, remainder = n % k;
320
ListNode[] res = new ListNode[k];
4-
int n = getLength(root);
5-
int len = n / k, left = n % k;
6-
ListNode pre = null; // 记录链尾
7-
for (int i = 0; i < k && root != null; ++i) {
8-
res[i] = root;
9-
int step = len;
10-
if (left > 0) {
11-
--left;
12-
++step;
21+
cur = root;
22+
for (int i = 0; i < k; ++i) {
23+
ListNode head = cur;
24+
for (int j = 0; j < width + ((i < remainder) ? 1 : 0) - 1; ++j) {
25+
if (cur != null) {
26+
cur = cur.next;
27+
}
1328
}
14-
for (int j = 0; j < step; ++j) {
15-
pre = root;
16-
root = root.next;
29+
if (cur != null) {
30+
ListNode t = cur.next;
31+
cur.next = null;
32+
cur = t;
1733
}
18-
pre.next = null; // 断链
19-
}
20-
return res;
21-
}
22-
23-
private int getLength(ListNode root) {
24-
int res = 0;
25-
while (root != null) {
26-
++res;
27-
root = root.next;
34+
res[i] = head;
2835
}
2936
return res;
3037
}
31-
}
38+
}

‎solution/0700-0799/0725.Split Linked List in Parts/Solution.py

+24
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,24 @@
1+
# Definition for singly-linked list.
2+
# class ListNode:
3+
# def __init__(self, x):
4+
# self.val = x
5+
# self.next = None
6+
7+
class Solution:
8+
def splitListToParts(self, root: ListNode, k: int) -> List[ListNode]:
9+
n, cur = 0, root
10+
while cur:
11+
n += 1
12+
cur = cur.next
13+
cur = root
14+
width, remainder = divmod(n, k)
15+
res = [None for _ in range(k)]
16+
for i in range(k):
17+
head = cur
18+
for j in range(width + (i < remainder) - 1):
19+
if cur:
20+
cur = cur.next
21+
if cur:
22+
cur.next, cur = None, cur.next
23+
res[i] = head
24+
return res

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