### Insertion_Sort_List

Sort a linked list using insertion sort.

A graphical example of insertion sort. The partial sorted list (black) initially contains only the first element in the list. With each iteration one element (red) is removed from the input data and inserted in-place into the sorted list.

Algorithm of Insertion Sort:

1. Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list.
2. At each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there.
3. It repeats until no input elements remain.

Example 1:

Input: 4->2->1->3Output: 1->2->3->4

Example 2:

Input: -1->5->3->4->0Output: -1->0->3->4->5

### Analyze

head0: 当前已排序列表的最后一个; pre: 用于遍历当前已排序列表;

head0dummy -> 4 -> 2 -> 1 -> 3            .            .pre    head0dummy -> 4 -> 2 -> 1 -> 3            .            .pre         head0dummy -> 2 -> 4 -> 1 -> 3            .            .             pre head0dummy -> 1 -> 2 -> 4 -> 3            .            .                      head0dummy -> 1 -> 2 -> 3 -> 4
head0dummy -> -1 -> 5 -> 3 -> 4 -> 0              .              .         pre head0dummy -> -1 -> 5 -> 3 -> 4 -> 0              .              .              pre head0dummy -> -1 -> 3 -> 5 -> 4 -> 0              .              .              pre head0dummy -> -1 -> 3 -> 5 -> 4 -> 0              .              .         pre           head0dummy -> -1 -> 3 -> 4 -> 5 -> 0              .              .                            head0dummy -> -1 -> 0 -> 3 -> 4 -> 5
/** * Definition for singly-linked list. * function ListNode(val) { *     this.val = val; *     this.next = null; * } *//** * @param {ListNode} head * @return {ListNode} */var insertionSortList = function(head) {  const dummy = new ListNode(0)  dummy.next = head  let head0 = dummy.next
while (head0 && head0.next) {    if (head0.next.val >= head0.val) {      head0 = head0.next      continue    }
let pre = dummy    while (pre.next.val < head0.next.val) { pre = pre.next }
let next = head0.next    head0.next = next.next    next.next = pre.next    pre.next = next  }
return dummy.next}

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