23.合并K个排序链表

23.合并K个排序链表

1.逐步合并

会两个链表合并，给的list中的许多链表遍历一遍，逐步合并

合并两个

public ListNode mergeTwoLists(ListNode a, ListNode b) {
    if (a == null || b == null) {
        return a != null ? a : b;
    }
    ListNode head = new ListNode(0);
    ListNode tail = head, aPtr = a, bPtr = b;
    while (aPtr != null && bPtr != null) {
        if (aPtr.val < bPtr.val) {
            tail.next = aPtr;
            aPtr = aPtr.next;
        } else {
            tail.next = bPtr;
            bPtr = bPtr.next;
        }
        tail = tail.next;
    }
    tail.next = (aPtr != null ? aPtr : bPtr);
    return head.next;
}

逐步合并

class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        ListNode ans = null;
        for (int i = 0; i < lists.length; ++i) {
            ans = mergeTwoLists(ans, lists[i]);
        }
        return ans;
    }

    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null || b == null) {
            return a != null ? a : b;
        }
        ListNode head = new ListNode(0);
        ListNode tail = head, aPtr = a, bPtr = b;
        while (aPtr != null && bPtr != null) {
            if (aPtr.val < bPtr.val) {
                tail.next = aPtr;
                aPtr = aPtr.next;
            } else {
                tail.next = bPtr;
                bPtr = bPtr.next;
            }
            tail = tail.next;
        }
        tail.next = (aPtr != null ? aPtr : bPtr);
        return head.next;
    }
}

方法二：分治合并

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];
        }
        if (l > r) {
            return null;
        }
        int mid = (l + r) >> 1;
        return mergeTwoLists(merge(lists, l, mid), merge(lists, mid + 1, r));
    }

    public ListNode mergeTwoLists(ListNode a, ListNode b) {
        if (a == null || b == null) {
            return a != null ? a : b;
        }
        ListNode head = new ListNode(0);
        ListNode tail = head, aPtr = a, bPtr = b;
        while (aPtr != null && bPtr != null) {
            if (aPtr.val < bPtr.val) {
                tail.next = aPtr;
                aPtr = aPtr.next;
            } else {
                tail.next = bPtr;
                bPtr = bPtr.next;
            }
            tail = tail.next;
        }
        tail.next = (aPtr != null ? aPtr : bPtr);
        return head.next;
    }
}

作者：力扣官方题解
链接：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/
来源：力扣（LeetCode）

参考转载：https://leetcode.cn/problems/merge-k-sorted-lists/solutions/219756/he-bing-kge-pai-xu-lian-biao-by-leetcode-solutio-2/