标签：返回 == ide nod 示例 排序 ati 复杂度 divide
合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。
示例:
输入: [ 1->4->5, 1->3->4, 2->6 ] 输出: 1->1->2->3->4->4->5->6
解法一：
暴力法：逐一合并，直到数组结尾
public ListNode mergeKLists(ListNode[] lists) { int len = lists.length; if(len == 0) return null; int len = lists.length; if(len == 0) return null; ListNode dummy = new ListNode(-1); dummy.next = lists[0]; int i = 1; while(i<len){ dummy.next = merge(dummy.next,lists[i]); i = i + 1; } return dummy.next; } //合并两个链表 public static ListNode merge(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode node = dummyHead; while(l1!=null && l2!=null){ if(l1.val>=l2.val){ node.next = l2; l2 = l2.next; }else{ node.next = l1; l1 = l1.next; } node = node.next; } if(l1==null){ node.next = l2; } if(l2==null){ node.next = l1; } return dummyHead.next; }
解法二：
归并思想;
public ListNode mergeKLists(ListNode[] lists) { int len = lists.length; if(len == 0) return null; return divide(lists,0,lists.length-1); } public static ListNode divide(ListNode[] lists,int l,int r){ if(r == l){ return lists[l]; } int mid = (r - l) / 2 + l; ListNode l1 = divide(lists,l,mid); ListNode l2 = divide(lists,mid+1,r); return merge(l1,l2); } //合并两个链表 public static ListNode merge(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode node = dummyHead; while(l1!=null && l2!=null){ if(l1.val>=l2.val){ node.next = l2; l2 = l2.next; }else{ node.next = l1; l1 = l1.next; } node = node.next; } if(l1==null){ node.next = l2; } if(l2==null){ node.next = l1; } return dummyHead.next; }
//合并两个链表 public static ListNode merge(ListNode l1, ListNode l2) { ListNode dummyHead = new ListNode(0); ListNode node = dummyHead; while(l1!=null && l2!=null){ if(l1.val>=l2.val){ node.next = l2; l2 = l2.next; }else{ node.next = l1; l1 = l1.next; } node = node.next; }
if(l1==null){ node.next = l2; } if(l2==null){ node.next = l1; } return dummyHead.next; }
标签：返回 == ide nod 示例 排序 ati 复杂度 divide
原文地址：https://www.cnblogs.com/du001011/p/10668165.html
