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Merge Sort Array: 看完stanford的 CS106B的video，https://www.youtube.com/watch?v=LlNawf0JeF0&list=PLnfg8b9vdpLn9exZweTJx44CII1bYczuk&index=55 醍醐灌顶；
public class Solution {/*** @param A: an integer array* @return: nothing*/public void sortIntegers2(int[] A) {// merge sort;if(A == null || A.length == 0) return;int[] temp = new int[A.length];mergeSort(A, 0, A.length-1, temp);}private void mergeSort(int[] A, int start, int end, int[] temp) {if(start >= end) {return;}int mid = start + (end - start) / 2;mergeSort(A, start, mid, temp);mergeSort(A, mid+1, end, temp);merge(A, start, end, temp);}private void merge(int[] A, int start, int end, int[] temp) {int mid = start + (end - start) / 2;int astart = start;int bstart = mid+1;int index = astart; // NOTE: index is astart;while(astart <= mid && bstart <= end) {if(A[astart] < A[bstart]) {temp[index++] = A[astart++];} else { // A[bstart] < A[astart];temp[index++] = A[bstart++];}}while(astart <= mid) {temp[index++] = A[astart++];}while(bstart <= end) {temp[index++] = A[bstart++];}// give temp to A;for(int i=start; i<=end; i++){A[i] = temp[i];}}
}Merge Sort LinkedList:
Sort a linked list in O(n log n) time using constant space complexity.
思路：merge sort 是nlogn
注意：
1. 分开两个list，找中点的时候，一般是找中点的前一个点，因为前面的list最后一个需要变成null，跟后面分开。
2. merge的时候，为了返回head node必须有个dump，hold住head，然后有个cur作为指针，连接下一个node。
/*** Definition for singly-linked list.* public class ListNode {* int val;* ListNode next;* ListNode(int x) { val = x; }* }*/
class Solution {private ListNode findMiddle(ListNode node) {ListNode dummpy = new ListNode(0);dummpy.next = node;ListNode slow = dummpy;ListNode fast = dummpy;while(fast != null && fast.next != null) {slow = slow.next;fast = fast.next.next;}return slow;}public ListNode sortList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode slow = findMiddle(head);ListNode newhead = slow.next;slow.next = null;ListNode l1 = sortList(head);ListNode l2 = sortList(newhead);return MergeTwoSortedList(l1, l2);}private ListNode MergeTwoSortedList(ListNode l1, ListNode l2) {ListNode dummpy = new ListNode(0);ListNode cur = dummpy;while(l1 != null && l2 != null) {if(l1.val < l2.val) {cur.next = l1;l1 = l1.next;cur = cur.next;} else {cur.next = l2;l2 = l2.next;cur = cur.next;}}if(l1 != null) {cur.next = l1;}if(l2 != null) {cur.next = l2;}return dummpy.next;}
}3. 这题再联想到：如何用 insertion sort 去sort integer array和 linked list
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