leetcode 23 合併k個排序鏈表

方法1 ：暴力法

用一個數組將所有的鏈表節點放入數組中然後進行直接排序

	/**
     * 暴力法
     * @param lists
     * @return
     */
    public ListNode mergeKLists1(ListNode[] lists){
        int n = 0;
        ListNode ans = new ListNode(0);
        ArrayList<Integer> arrayList = new ArrayList<>();
        for (ListNode list : lists) {
            while (list.next != null){
                arrayList.add(list.val);
                list = list.next;
            }
        }
        arrayList.sort(new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o1.compareTo(o2);
            }
        });
        ListNode temp = ans;
        for (Integer integer : arrayList) {
            ans.val = integer;
            ans = ans.next;
        }
        return ans;
    }

方法2：歸併思想

每次合併兩個鏈表直到只有一個鏈表爲止。

圖片來自力扣中國leetcode題解

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Main {
    public ListNode mergeKLists(ListNode[] lists) {
        int len = lists.length;
        if(len == 0) return null;
        return merge(lists,0,lists.length-1);
    }

    public static ListNode merge(ListNode[] lists,int l,int r){
        if(r == l){
            return lists[l];
        }
        int mid = (r - l) / 2 + l;
        ListNode l1 = merge(lists,l,mid);
        ListNode l2 = merge(lists,mid+1,r);
        return merge(l1,l2);
    }
    //合併兩個鏈表
    public static ListNode merge(ListNode l1, ListNode l2) {
        ListNode ansHead = new ListNode(0);
        ListNode node = ansHead;
        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 ansHead.next;
    }
}