【合并排序链表】分治/优先队列

合并两个排序链表

• 模拟维护一个合并链表，每次添加两个排序链表中较小val的节点即可

模拟代码

    public ListNode mergeTwo(ListNode a, ListNode b) {
        if(a == null) return b;
        if(b == null) return a;
        ListNode ans = new ListNode(0);
        ListNode ans0 = ans;
        while(a != null && b != null) {
            if(a.val <= b.val) {
                ans.next = new ListNode(a.val, null);
                a = a.next;
            } else {
                ans.next = new ListNode(b.val, null);
                b = b.next;
            }
            ans = ans.next;
        }
        if(a != null) {
            ans.next = a;
        } else {
            ans.next = b;
        }
        return ans0.next;
    }

合并K个排序链表

leetcode 23. 合并 K 个升序链表

顺序合并

• 两两合并链表即可

顺序合并代码

/**
 * 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; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        int len = lists.length;
        ListNode ans = null;
        for(int i = 0; i < len; ++ i ) {
            ans = mergeTwo(ans, lists[i]);
        }
        return ans;
    }
 
    public ListNode mergeTwo(ListNode a, ListNode b) {
        if(a == null) return b;
        if(b == null) return a;
        ListNode ans = new ListNode(0);
        ListNode ans0 = ans;
        while(a != null && b != null) {
            if(a.val <= b.val) {
                ans.next = new ListNode(a.val, null);
                a = a.next;
            } else {
                ans.next = new ListNode(b.val, null);
                b = b.next;
            }
            ans = ans.next;
        }
        if(a != null) {
            ans.next = a;
        } else {
            ans.next = b;
        }
        return ans0.next;
    }
}

分治合并

• 基于两两合并，再分治思想递归合并链表数组

分治合并代码

/**
 * 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; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        return merge(lists, 0, lists.length - 1);
    }
 
    public ListNode merge(ListNode[] lists, int l, int r) {
        if(l == r) return lists[l];
        // 边界值：lists.length = 0会出现 l>r
        if(l > r) return null;
        int mid = l + ((r - l) >> 1);
        
        return mergeTwo(merge(lists, l, mid), merge(lists, mid + 1, r));
    }
 
    public ListNode mergeTwo(ListNode a, ListNode b) {
        if(a == null) return b;
        if(b == null) return a;
        ListNode ans = new ListNode(0);
        ListNode ans0 = ans;
        while(a != null && b != null) {
            if(a.val <= b.val) {
                ans.next = new ListNode(a.val, null);
                a = a.next;
            } else {
                ans.next = new ListNode(b.val, null);
                b = b.next;
            }
            ans = ans.next;
        }
        if(a != null) {
            ans.next = a;
        } else {
            ans.next = b;
        }
        return ans0.next;
    }
}

使用优先队列合并

使用优先队列维护两个排序链表中的值，然后按照优先队列中升序元素构造新的合并链表即可

使用优先队列合并代码

/**
 * 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; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        int len = lists.length;
        Queue<Integer> pq = new PriorityQueue<>();
        for(int i = 0; i < len; ++ i ) {
            while(lists[i] != null) {
                pq.offer(lists[i].val);
                lists[i] = lists[i].next;
            }
        }
        ListNode head, tail;
        head = tail = null;
        while(!pq.isEmpty()) {
            int val = pq.poll();
            if(head == null) head = tail = new ListNode(val, null);
            else {
                tail.next = new ListNode(val, null);
                tail = tail.next;
            }
        }
        return head;
    }
}