解题思路:
找到链表中点: 使用快慢指针法,快指针每次移动两步,慢指针每次移动一步。当快指针到达末尾时,慢指针指向中点。递归分割与排序: 将链表从中点处分割为左右两个子链表,分别对这两个子链表递归排序。合并有序子链表: 将两个已排序的子链表合并成一个有序链表。
Java代码:
class Solution { public ListNode sortList ( ListNode head) { if ( head == null || head. next == null ) { return head; } ListNode slow = head, fast = head; while ( fast. next != null && fast. next. next != null ) { slow = slow. next; fast = fast. next. next; } ListNode mid = slow. next; slow. next = null ; ListNode left = sortList ( head) ; ListNode right = sortList ( mid) ; return merge ( left, right) ; } private ListNode merge ( ListNode l1, ListNode l2) { ListNode dummy = new ListNode ( - 1 ) ; ListNode current = dummy; while ( l1 != null && l2 != null ) { if ( l1. val <= l2. val) { current. next = l1; l1 = l1. next; } else { current. next = l2; l2 = l2. next; } current = current. next; } current. next = ( l1 != null ) ? l1 : l2; return dummy. next; }
}
复杂度分析:
时间复杂度: 归并排序的时间复杂度为 O(nlogn)。空间复杂度: O(log n)。(递归栈深度)
解题思路:
初始化优先队列: 将所有链表的头节点加入堆中,堆顶元素为当前最小值。构建结果链表: 每次从堆顶取出最小节点,添加到结果链表中,并将其下一个节点加入堆中(若存在)。处理空链表: 跳过输入数组中的空链表,避免无效操作。
Java代码:
class Solution { public ListNode mergeKLists ( ListNode [ ] lists) { if ( lists == null || lists. length == 0 ) return null ; PriorityQueue < ListNode > minHeap = new PriorityQueue < > ( ( a, b) -> a. val - b. val) ; for ( ListNode node : lists) { if ( node != null ) { minHeap. offer ( node) ; } } ListNode dummy = new ListNode ( - 1 ) ; ListNode current = dummy; while ( ! minHeap. isEmpty ( ) ) { ListNode smallest = minHeap. poll ( ) ; current. next = smallest; current = current. next; if ( smallest. next != null ) { minHeap. offer ( smallest. next) ; } } return dummy. next; }
}
复杂度分析:
时间复杂度: O(nklogk),其中 n 是总节点数,k 是链表数量。空间复杂度: O(k),用于存储堆中的节点。