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出处 LeetCode 算法第148题 在 O(n log n) 时间复杂度和常数级空间复杂度下,对链表进行排序。 示例 1: 输入: 4->2->1->3 输出: 1->2->3->4 示例 2: 输入: -1->5->3->4->0 输出: -1->0->3->4->5
出处 LeetCode 算法第148题
在 O(n log n) 时间复杂度和常数级空间复杂度下,对链表进行排序。
示例 1:
输入: 4->2->1->3 输出: 1->2->3->4
示例 2:
输入: -1->5->3->4->0 输出: -1->0->3->4->5
排序算法有很多种,此题要求时间复杂度在 O(n log n),而且是链表的排序,所以采用归并排序是比较合适的。具体思路为:找到链表的中间节点,将左右两边的链表各自进行排序,最后进行归并,将两个排好序的链表合并为一个单独的链表(左右链表各自进行递归的排序操作)
var sortList = function (head) { if (!head || !head.next) { return head; } var start = head; var count = 0; while (start) { count++; start = start.next; } var middle = Math.floor(count / 2); var prev = head; var current = head.next; for (var i = 0; i < middle - 1; i++) { prev = prev.next; current = current.next; } var left = head; prev.next = null; var right = current; left = sortList(left); right = sortList(right) var result = merge(left, right); return result; }; function merge(left, right) { var temp = new ListNode(); var _start = temp; while (left && right) { if (left.val <= right.val) { temp.next = left; left = left.next; } else { temp.next = right; right = right.next; } temp = temp.next; } if (left) { temp.next = left; } if (right) { temp.next = right } return _start.next; }
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习题
思路
排序算法有很多种,此题要求时间复杂度在 O(n log n),而且是链表的排序,所以采用归并排序是比较合适的。具体思路为:找到链表的中间节点,将左右两边的链表各自进行排序,最后进行归并,将两个排好序的链表合并为一个单独的链表(左右链表各自进行递归的排序操作)
解答
The text was updated successfully, but these errors were encountered: