标签:art complex sort list else list val java ret blog
Sort a linked list in O(n log n) time using constant space complexity.
排序,要求是O(nlog(n))的时间复杂度和常数的空间复杂度,那么就使用归并就可以了。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode sortList(ListNode head) { if( head == null || head.next == null) return head; int size = 1; ListNode start = new ListNode(0); start.next = head; while( true ){ ListNode node1 = start; ListNode node2 = start.next; for( int i = 0 ; i < size && node2!=null;i++){ node2 = node2.next; } if( node2 == null ) break; ListNode nnn = start.next; while( node2 != null ){ node1 = helper(node1,node2,size); if( node1 == null ) break; node2 = node1.next; for( int i = 0 ; i< size && node2 != null;i++){ node2 = node2.next; } } size*=2; } return start.next; } public ListNode helper(ListNode node1,ListNode node2,int size){ int num1 = 0,num2 = 0; ListNode node = null; if( node1.next.val < node2.val ){ node = node1.next; node1 = node1.next.next; num1++; }else{ ListNode nn = node1.next; node1.next = node2; node1 = nn; node = node2; node2 = node2.next; num2++; } while( num1 < size && num2 < size && node1 != null && node2 != null){ if( node1.val < node2.val ){ node.next = node1; node = node1; node1 = node1.next; num1++; }else{ node.next = node2; node = node2; node2 = node2.next; num2++; } } while( num1 < size && node1 != null){ node.next = node1; node = node1; node1 = node1.next; num1++; } while( num2 < size && node2 != null){ node.next = node2; node = node2; node2 = node2.next; num2++; } node.next = node2; return node; } }
leetcode 148. Sort List ----- java
标签:art complex sort list else list val java ret blog
原文地址:http://www.cnblogs.com/xiaoba1203/p/6076961.html