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; }