[leetCode刷题笔记]2017.02.01

最新推荐文章于 2021-08-06 15:33:05 发布

原创最新推荐文章于 2021-08-06 15:33:05 发布 · 403 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetCode

本文介绍了几种链表操作的方法，包括链表的翻转、旋转、分区及排序等。通过具体的代码实现展示了如何解决这些常见问题。

25. Reverse Nodes in k-Group

这道题有点小复杂。

先求出list的长度。再在while-loop中套一个for-loop，for-loop中将k个node进行reverse，然后再移动到下组

public class Solution {

public ListNode reverseKGroup(ListNode head, int k) {
ListNode prevNode = new ListNode(0);
prevNode.next = head;
ListNode curr = prevNode;
// get the list length
int i = count(head);
while (i >= k) {
// reverse k node in the list
for (int j = 0; j < k - 1; j++) {
ListNode nextNode = curr.next;
curr.next = head.next;
head.next = curr.next.next;
curr.next.next = nextNode;
}
curr = head;
head = head.next;
i -= k;
}
return prevNode.next;
}
private int count(ListNode head) {
ListNode curr = head;
int i= 0;
while (curr != null) {
i++;
curr = curr.next;
}
return i;
}

}

61. Rotate List

这道题先获得尾部node，再将尾部node的next连上head形成一个环。k可能大于list长度n，所以获得断开点的位置是n - k % n,断开后再返回指针即可。

public class Solution {
public ListNode rotateRight(ListNode head, int k) {
if (head == null) {
return head;
}
ListNode tail = head;
int length = count(head);
// get the tail of List
while (tail.next != null) {
tail = tail.next;
}
// generate the circle
tail.next = head;
ListNode breakNode = tail;

for (int i = 0; i < (length - k % length); i++) {
breakNode = breakNode.next;
}
ListNode output = breakNode.next;
// break the circle
breakNode.next = null;
return output;
}
private int count(ListNode head) {
ListNode curr = head;
int i= 0;
while (curr != null) {
i++;
curr = curr.next;
}
return i;

}
}

86. Partition List

将list按X分成两段,小的的一段大的一段，再连起来。。。

public class Solution {
public ListNode partition(ListNode head, int x) {
ListNode big = new ListNode(0);
ListNode bigCurr = big;
ListNode small = new ListNode(0);
ListNode smallCurr = small;
//遍历
while(head != null){
if(head.val < x){//<x成一组
smallCurr.next = head;
smallCurr = smallCurr.next;
}
else {//>=x成一组
bigCurr.next = head;
bigCurr = bigCurr.next;
}
head = head.next;
}
smallCurr.next = big.next;
bigCurr.next = null;//斩断后面的连接
return small.next;
}
}

148. Sort List

重复使用merge two sorted list方法，将list不断分成两部分，分别进行排序。mergesort思想。。。

public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}

ListNode fast = head;
ListNode slow = head;
while (fast.next != null) {
fast = fast.next.next;
if (fast == null) {
break;
}

slow = slow.next;
}
// slow node current in center
ListNode right = slow.next;
slow.next = null;

ListNode left = sortList(head);
right = sortList(right);

return mergeTwoLists(left, right);
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode lastNode = dummy;

while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
lastNode.next = l1;
l1 = l1.next;
} else {
lastNode.next = l2;
l2 = l2.next;
}
lastNode = lastNode.next;
}

if (l1 != null) {
lastNode.next = l1;
} else {
lastNode.next = l2;
}

return dummy.next;
}
}