148. Sort List

这题就是merge sort的链表实现。
先看一下mergeSort:

public class myMergeSort {
    static int number = 0;

    public static void main(String[] args) {
        int[] a = {26, 5, 98, 108, 28, 99, 100, 56, 34, 1, 45, 69};
        printArray("排序前：", a);
        MergeSort(a);
        printArray("排序后：", a);
    }

    private static void printArray(String pre, int[] a) {
        System.out.print(pre + "\n");
        for (int i = 0; i < a.length; i++)
            System.out.print(a[i] + "\t");
        System.out.println();
    }

    private static void MergeSort(int[] a) {
        // TODO Auto-generated method stub
        System.out.println("开始排序");
        Sort(a, 0, a.length - 1);
    }

    private static void Sort(int[] a, int left, int right) {
        if (left >= right)
            return;
        int mid = (left + right) / 2;
        //二路归并排序里面有两个Sort，多路归并排序里面写多个Sort就可以了
        Sort(a, left, mid);
        Sort(a, mid + 1, right);
        merge(a, left, mid, right);
    }

    private static void merge(int[] a, int left, int mid, int right) {
        int[] tmp = new int[a.length];
        int r1 = mid + 1;
        int tIndex = left;
        int cIndex = left;
        // 逐个归并
        while (left <= mid && r1 <= right) {
            if (a[left] <= a[r1])
                tmp[tIndex++] = a[left++];
            else
                tmp[tIndex++] = a[r1++];
        }
        // 将左边剩余的归并
        while (left <= mid) {
            tmp[tIndex++] = a[left++];
        }
        // 将右边剩余的归并
        while (r1 <= right) {
            tmp[tIndex++] = a[r1++];
        }
        System.out.println("第" + (++number) + "趟排序:\t");
        // TODO Auto-generated method stub
        //从临时数组拷贝到原数组
        while (cIndex <= right) {
            a[cIndex] = tmp[cIndex];
            //输出中间归并排序结果
            System.out.print(a[cIndex] + "\t");
            cIndex++;
        }
        System.out.println();
    }
}

复杂度O(nlogn)。
这题的解法我看了https://leetcode.com/problems/sort-list/#/solutions
前半部分divide的递归部分比较难，后半部分就是merge two sorted lists。

    public ListNode sortList(ListNode head) {

        if (head == null || head.next == null) {
            return head;
        }
        ListNode prev = new ListNode(-1);
        ListNode slow = head;
        ListNode fast = head;
        while (fast != null && fast.next != null) {
            prev = slow;
            slow = slow.next;
            fast = fast.next.next;
        }
        //prev把slow前面那个node跟slow之间的link断开
        prev.next = null;
        ListNode l1 = sortList(head);
        ListNode l2 = sortList(slow);
        return merge(l1, l2);
    }

    private ListNode merge(ListNode l1, ListNode l2) {
        ListNode node = new ListNode(-1);
        ListNode l = node;
        while (l1 != null && l2 != null) {
            if (l1.val < l2.val) {
                l.next = l1;
                l1 = l1.next;
            } else {
                l.next = l2;
                l2 = l2.next;
            }
            l = l.next;
        }
        if (l1 != null) {
            l.next = l1;
        }
        if (l2 != null) {
            l.next = l2;
        }
        return node.next;
    }