🌗 23. 合并K个升序链表

难度: 🌗

问题描述

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解法

/**
 * 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) {
        // 思路:
        // 先分治，后两两合并
        int len = lists.length;
        // 特殊情况特判
        if(len == 0) {
            return null;
        }
        if(len == 1) {
            return lists[0];
        }
        // len > 1
        return mySol(lists, 0, len - 1);
    }

    private ListNode mySol(ListNode[] lists, int left, int right) {
        // 递归终止条件
        if(left >= right) {
            return lists[left];
        }
        // left < right
        int mid = left + ((right - left) >> 1);
        ListNode a = mySol(lists, left, mid);
        ListNode b = mySol(lists, mid + 1, right);
        return merge(a, b);
    }

    private ListNode merge(ListNode a, ListNode b) {
        if(a == null) {
            return b;
        }
        if(b == null) {
            return a;
        }
        ListNode res = new ListNode(-1);
        ListNode cur = res;
        while(a != null && b != null) {
            if(a.val <= b.val) {
                cur.next = a;
                cur = cur.next;
                a = a.next;
            } else {
                cur.next = b;
                cur = cur.next;
                b = b.next;
            }
        }
        while(a != null) {
            cur.next = a;
            a = a.next;
            cur = cur.next;
        }
        while(b != null) {
            cur.next = b;
            cur = cur.next;
            b = b.next;
        }
        return res.next;
    }
}

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输出