Sort List[leetcode] 由归并排序的递归和循环，到本题的两种解法

归并排序能够有两种思路----top-down 和 bottom-up

top-down:

递归实现，将数组分成两半。分别处理。再合并。

伪代码例如以下：

split ( A[], l, r)
{if ( r - l < 2) return;m = (r + l) / 2;split ( A, l, m); //split A[l…m-1]split ( A, m, r); //split A[m…r-1]merge ( A, l, m, e); //merge A[l…m-1] and A[m…e-1]
}

bottom-up:

循环实现。将数组看做n个长度为1的组数组。

用width控制每次merge的子数组长度。width每次翻倍

伪代码例如以下：

sort ( A[], n)
{for (width = 1; width < n; width *= 2){for (i = 0; i < n; i += 2 * width){merge(A, i, min(i + width, n), min(i + 2 * width, n));}}
}

Sort list中使用链表。不能在O(1)的时间内訪问随意节点，同一时候注意要处理尾部节点的next，置为NULL

和上面的伪代码类似，首先实现merge函数：

    ListNode * merge(ListNode * h1, int s1, ListNode * h2, int s2){if (h2 == NULL) return h1;ListNode * h;if (h1->val < h2->val)h = advance(h1, s1);elseh = advance(h2, s2);ListNode * cur = h;while (s1 && s2){if (h1->val < h2->val)cur->next = advance(h1, s1);elsecur->next = advance(h2, s2);cur = cur->next;}if (s1){cur->next = h1;while(s1) advance(h1, s1);}if (s2){cur->next = h2;while(s2) advance(h2, s2);}return h;}ListNode * advance(ListNode * (& n), int & size){ListNode * temp = n;if (size == 1)  n->next = NULL;n = n->next;size--;return temp;}

同一时候实现工具函数，訪问任何位置节点

ListNode * getNode(ListNode * head, int len){while (len -- && head) head = head->next;return head;}

循环版本号主函数例如以下：

ListNode *sortList(ListNode *head) {ListNode * cur = head;int size = 0;while (cur){size ++;cur = cur->next;}ListNode * pre;for (int w = 1; w <= size; w *= 2){cur = head;for (int i = 0; i < size; i+= w*2){ListNode * h1 = cur, * h2 = getNode(cur, w), * next = getNode(cur, 2 * w);cur = merge(h1, min(w, size - i), h2, min(w, size - i - w));if (i == 0)     head = cur;else            pre->next = cur;pre = getNode(cur, min(2 * w, size - i) - 1);cur = next;}}return head;}

递归版本号主函数例如以下：

    ListNode *sortList(ListNode *head) {ListNode * cur = head;int size = 0;while (cur){size ++;cur = cur->next;}return sort(head, size - size / 2, getNode(head, size - size / 2), size / 2);}ListNode * sort(ListNode * h1, int s1, ListNode * h2, int s2){if (s1 == 0) return h2;if (s2 == 0) return h1;h1 = sort(h1, s1 - s1 / 2, getNode(h1, s1 - s1 / 2), s1 / 2);h2 = sort(h2, s2 - s2 / 2, getNode(h2, s2 - s2 / 2), s2 / 2);return merge(h1, s1, h2, s2);}