Leetcode Sort List非递归

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Sort a linked list in O(n log n) time using constant space complexity.
利用MergeSort非递归方式完成该题
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/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *sortList(ListNode *head) {        int interval = 1;        while(1){            ListNode *node = head;            bool UpdateHead = true;            ListNode *preNode = NULL;            while(1){                ListNode *first_link_head, *first_link_tail, *second_link_head, *second_link_tail;                first_link_head=first_link_tail=second_link_head=second_link_tail=node;                int cnt = interval;                while(second_link_head && cnt--){                    first_link_tail = second_link_head;                     second_link_tail = second_link_head = second_link_head->next;                }                if(second_link_head){                    cnt = interval;                                        while(second_link_tail->next && --cnt){                        second_link_tail = second_link_tail->next;                     }                                        ListNode *merge_link, *merge_link_head;                    merge_link = merge_link_head = NULL;                    node = second_link_tail->next;                     second_link_tail->next = NULL;                     first_link_tail->next = NULL;                    while( second_link_head || first_link_head ){                       if(!second_link_head ||                            first_link_head && second_link_head &&                            first_link_head->val < second_link_head->val){                           merge_link = !merge_link ?                                          first_link_head :                                          merge_link->next = first_link_head;                           first_link_head = first_link_head->next;                       }else{                           merge_link = !merge_link ?                                          second_link_head :                                          merge_link->next = second_link_head;                           second_link_head = second_link_head->next;                       }                       if(merge_link_head == NULL){                          merge_link_head = merge_link;                         }                       if(UpdateHead){                           head = merge_link;                           UpdateHead = false;                       }                    }                                        if(preNode != NULL){                       preNode->next = merge_link_head;                    }                    preNode = merge_link;                }else{                    if(preNode != NULL){                       preNode->next = node;                    }                    break;                }                }            if(UpdateHead){                break;            }            interval *= 2;        }        return head;    }};
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