[leetcode]148. Sort List -- JavaScript代码

题目要求时间复杂度为O(nlogn)并且空间复杂度为O(1)：归并排序的时间复杂度合格，并且由于这是链表排序，因此，空间复杂度也可以做到O(1)。

关于归并排序，百度上有现成的代码，本题我采用的是递归的方式。不过这道题需要我们根据链表的特点做一些修改：例如，找中间节点的函数，需要使用快慢指针的方法，以完成高效的查找中间指针。

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var sortList = function(head) {// 归并排序的时间复杂度为O(nlogn) 
    if(head===null){
        return head;
    }
    ret = mergeSort(head);
    return ret;

    function mergeSort(items){// 归并算法
        if(items.next===null){
            return items;
        }
        var left = items;
        var right = getMid(items);
        return　merge(mergeSort(left), mergeSort(right));
    }

    function merge(left,right){// 合并
        var merge_result;
        if(left===null){
            return right;
        }else if(right===null){
            return left;
        }
        if(left.val<right.val){
            lead = left;
            left = left.next;
        }else{
            lead = right;
            right = right.next;
        }
        var tmp = lead;
        while(left!==null && right!==null){
            if(left.val<right.val){
                tmp.next = left;
                tmp = left;
                left = left.next;
            }else{
                tmp.next = right;
                tmp = right;
                right = right.next;
            }
        }
        if(left!==null){
            tmp.next = left;
        }else{
            tmp.next = right;
        }
        return lead;
    }

    function getMid(first){// 快慢指针获得中间点
        //guaranteed that at least two nodes
        var fast = first.next;
        var slow = first.next;
        var prev = first;
        while(true)
        {
            if(fast !== null)
                fast = fast.next;
            else
                break;
            if(fast !== null)
                fast = fast.next;
            else
                break;
            prev = slow;
            slow = slow.next;
        }
        prev.next = null;  // cut
        return slow;
    }
};