[刷题]Merge k Sorted Lists

最新推荐文章于 2024-04-03 21:24:25 发布

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本文介绍两种合并K个已排序链表的方法：一种通过遍历寻找最小元素，另一种使用优先队列实现堆排序。这两种方法均实现了链表的合并，并返回一个排序后的链表。

[LintCode]Merge k Sorted Lists

Version I

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
    public ListNode mergeKLists(List<ListNode> lists) {  
        // 2015-08-21
        if (lists == null || lists.size() == 0) {
            return null;
        }
        
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy; // sorted list的尾节点
        // 这个数组保存了各链表中未排序的部分的表头
        ArrayList<ListNode> headList = new ArrayList<>(lists); 
        
        while (true) {
            int minValue = Integer.MAX_VALUE;
            int indexOfMin = -1;
            // 遍历数组，找到值最小的节点，并保存其序号
            for (int i = 0; i < headList.size(); i++) {
                ListNode temp = headList.get(i);
                if (temp == null) {
                    continue;
                }
                if (minValue > temp.val) {
                    minValue = temp.val;
                    indexOfMin = i;
                }                
            }
            // 更新sorted list和headList
            if (indexOfMin == -1) {
                break;
            } else {
                ListNode minNode = headList.get(indexOfMin);
                tail.next = minNode;
                tail = tail.next;
                headList.set(indexOfMin, minNode.next);
            }
        }
        return dummy.next;
    }
}

Version II 堆排序

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param lists: a list of ListNode
     * @return: The head of one sorted list.
     */
    private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
        public int compare(ListNode left, ListNode right) {
            if (left == null) {
                return 1;
            } else if (right == null) {
                return -1;
            }
            return left.val - right.val;
        }
    };
    
    public ListNode mergeKLists(List<ListNode> lists) {
        if (lists == null || lists.size() == 0) {
            return null;
        }
        
        Queue<ListNode> heap = new PriorityQueue<ListNode>(lists.size(), ListNodeComparator);
        for (int i = 0; i < lists.size(); i++) {
            if (lists.get(i) != null) {
                heap.add(lists.get(i));
            }
        }
        
        ListNode dummy = new ListNode(0);
        ListNode tail = dummy;
        while (!heap.isEmpty()) {
            ListNode head = heap.poll();
            tail.next = head;
            tail = head;
            if (head.next != null) {
                heap.add(head.next);
            }
        }
        return dummy.next;
    }
}