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87 lines (78 loc) · 2.56 KB
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// 23. 合并K个升序链表
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
/*
顺序合并：遍历链表数组，逐个链表进行合并
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode root = null;
for (ListNode head : lists) {
root = mergeTwoLists(root, head);
}
return root;
}
// 21.合并两个有序链表
private ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode head = new ListNode(-1);
ListNode pre = head;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
pre.next = list1;
list1 = list1.next;
} else {
pre.next = list2;
list2 = list2.next;
}
pre = pre.next;
}
pre.next = list1 == null ? list2 : list1;
return head.next;
}
}
/*
分治合并：
1、终止条件：左右边界相同则返回对应链表，左边界大于右边界则返回空
2、方法功能：入参链表数组、左边界、右边界，将链表数组拆分成两个链表，合并链表返回
3、递归逻辑：数组拆分后，左右两部分数组仍然需要继续拆分，直到获得两个链表进行合并，因此调用同个方法进行处理，合并后的链表返回给上一层继续合并
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
return merge(lists, 0, lists.length - 1);
}
private ListNode merge(ListNode[] lists, int left, int right) {
if (left == right) {
return lists[left];
}
if (left > right) {
return null;
}
int mid = (left + right) / 2;
return mergeTwoLists(merge(lists, left, mid), merge(lists, mid + 1, right));
}
// 21.合并两个有序链表
private ListNode mergeTwoLists(ListNode list1, ListNode list2) {
ListNode head = new ListNode(-1);
ListNode pre = head;
while (list1 != null && list2 != null) {
if (list1.val < list2.val) {
pre.next = list1;
list1 = list1.next;
} else {
pre.next = list2;
list2 = list2.next;
}
pre = pre.next;
}
pre.next = list1 == null ? list2 : list1;
return head.next;
}
}