标签:blog lis tac poi https -- ack 4sum csdn
2019/08/25: L230, L226, L18, L234, L23
1. L230 Kth smallest element in BST
借助 Stack 或 LinkedList inorder traveresal, LinkedList会更快
ArrayList, LinkedList, Vector, Stack https://blog.csdn.net/iteye_11495/article/details/82618748
2. L226 Invert Binary Tree
3. L18 4Sum
Two pointers
1 public List<List<Integer>> fourSum(int[] nums, int target) { 2 3 List<List<Integer>> res = new LinkedList<>(); 4 if(nums == null || nums.length < 4) return res; 5 Arrays.sort(nums); 6 int k, z; 7 for(int i = 0; i<nums.length-1;i++){ 8 if(i>0 && nums[i]==nums[i-1]) continue; 9 10 //pruning 11 if(nums[i]*4 > target) break; 12 if(nums[i] + 3* nums[nums.length-1]<target) continue; 13 14 for(int j =i+1; j<nums.length;j++){ 15 if(j>i+1 && nums[j]==nums[j-1]) continue; 16 17 //pruning 18 if(nums[i]+ nums[j]*3 >target) break; 19 if(nums[i] + nums[j]+ 2* nums[nums.length-1]<target) continue; 20 21 k = j+1; 22 z = nums.length -1; 23 while(k<z){ 24 int sum = nums[i] + nums[j] + nums[k] + nums[z]; 25 if(sum == target){ 26 List<Integer> list = new ArrayList<>(); 27 list.add(nums[i]); 28 list.add(nums[j]); 29 list.add(nums[k]); 30 list.add(nums[z]); 31 res.add(list); 32 while(k<z && nums[k]==nums[k+1])k++; 33 k++; 34 while(k<z && nums[z]==nums[z-1])z--; 35 z--; 36 }else if(sum > target) z--; 37 else k++; 38 } 39 } 40 } 41 return res; 42 }
加上pruning代码,beat 100%。
4. L234 Palindrome Linked List
slow fast pointers, 后半段reverse
Time O(n) spaceO(1)
public boolean isPalindrome(ListNode head) { if( head==null || head.next==null ) return true; ListNode slow = head; ListNode fast = head.next; while(fast!=null && fast.next!=null){ slow = slow.next; fast = fast.next.next; } ListNode pre = null; slow = slow.next; //reveres node after slow while(slow!=null){ ListNode next = slow.next; slow.next = pre; pre = slow; slow = next; } while(pre!=null && head!=null){ if(head.val!=pre.val) return false; head = head.next; pre = pre.next; } return true; }
5. L23. Merge k Sorted Lists
min heap
public ListNode mergeKLists(ListNode[] lists) { if(lists == null || lists.length == 0) return null; if(lists.length ==1) return lists[0]; PriorityQueue<ListNode> pq = new PriorityQueue<ListNode>(new Comparator<ListNode>(){ public int compare(ListNode l1, ListNode l2){ return l1.val - l2.val; } }); for(ListNode l : lists){ if(l!=null) pq.offer(l); } ListNode head = new ListNode(0); ListNode curr = head; while(!pq.isEmpty()){ ListNode newNode = pq.poll(); curr.next = newNode; curr = curr.next; if(newNode.next!=null){ pq.offer(newNode.next); } } return head.next; }
标签:blog lis tac poi https -- ack 4sum csdn
原文地址:https://www.cnblogs.com/rySZ/p/11411472.html