Reverse Linked List II leetcode java
2014-07-23 23:37
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题目:
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given
return
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
题解:
经典的题目就是链表逆序啦,一般的链表逆序是让把链表从前到后都逆序,这个是给定了起始位置和结束位置,方法是一样的。
就是维护3个指针,startpoint,node1和node2。
startpoint永远指向需要开始reverse的点的前一个位置。
node1指向正序中第一个需要rever的node,node2指向正序中第二个需要reverse的node。
交换后,node1 在后,node2在前。这样整个链表就逆序好了。
代码如下:
1 public ListNode reverseBetween(ListNode head, int m, int n) {
2 ListNode newhead = new ListNode(-1);
3 newhead.next = head;
4
5 if(head==null||head.next==null)
6 return newhead.next;
7
8 ListNode startpoint = newhead;//startpoint指向需要开始reverse的前一个
9 ListNode node1 = null;//需要reverse到后面去的节点
ListNode node2 = null;//需要reverse到前面去的节点
for (int i = 0; i < n; i++) {
if (i < m-1){
startpoint = startpoint.next;//找真正的startpoint
} else if (i == m-1) {//开始第一轮
node1 = startpoint.next;
node2 = node1.next;
}else {
node1.next = node2.next;//node1交换到node2的后面
node2.next = startpoint.next;//node2交换到最开始
startpoint.next = node2;//node2作为新的点
node2 = node1.next;//node2回归到node1的下一个,继续遍历
}
}
return newhead.next;
}
Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given
1->2->3->4->5->NULL, m = 2 and n = 4,
return
1->4->3->2->5->NULL.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
题解:
经典的题目就是链表逆序啦,一般的链表逆序是让把链表从前到后都逆序,这个是给定了起始位置和结束位置,方法是一样的。
就是维护3个指针,startpoint,node1和node2。
startpoint永远指向需要开始reverse的点的前一个位置。
node1指向正序中第一个需要rever的node,node2指向正序中第二个需要reverse的node。
交换后,node1 在后,node2在前。这样整个链表就逆序好了。
代码如下:
1 public ListNode reverseBetween(ListNode head, int m, int n) {
2 ListNode newhead = new ListNode(-1);
3 newhead.next = head;
4
5 if(head==null||head.next==null)
6 return newhead.next;
7
8 ListNode startpoint = newhead;//startpoint指向需要开始reverse的前一个
9 ListNode node1 = null;//需要reverse到后面去的节点
ListNode node2 = null;//需要reverse到前面去的节点
for (int i = 0; i < n; i++) {
if (i < m-1){
startpoint = startpoint.next;//找真正的startpoint
} else if (i == m-1) {//开始第一轮
node1 = startpoint.next;
node2 = node1.next;
}else {
node1.next = node2.next;//node1交换到node2的后面
node2.next = startpoint.next;//node2交换到最开始
startpoint.next = node2;//node2作为新的点
node2 = node1.next;//node2回归到node1的下一个,继续遍历
}
}
return newhead.next;
}
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