20190724算法题存档

题目描述

Sort a linked list in O(n log n) time using constant space complexity.

public class Solution {public ListNode sortList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode slow = head;ListNode quick = head;while(quick.next != null && quick.next.next != null) {slow = slow.next;quick = quick.next.next;}ListNode mid = slow;ListNode midNext = mid.next;mid.next = null;return merge(sortList(head), sortList(midNext));}private ListNode merge(ListNode node1, ListNode node2) {ListNode temp = new ListNode(0);ListNode cur = temp;while(node1 != null && node2 != null) {if(node1.val < node2.val) {cur.next = node1;node1 = node1.next;} else {cur.next = node2;node2 = node2.next;}cur = cur.next;}cur.next = node1 == null ? node2 : node1;return temp.next;}
}

题目描述

Sort a linked list using insertion sort.

public class Solution {public ListNode insertionSortList(ListNode head) {if(head == null || head.next == null) {return head;}ListNode temp = new ListNode(Integer.MIN_VALUE);ListNode cur = head;while(cur != null) {ListNode next = cur.next;ListNode pre = temp;while(pre.next != null && pre.next.val < cur.val ) {pre = pre.next;}cur.next = pre.next;pre.next = cur;cur = next;}return temp.next;}
}

题目描述

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree{1,#,2,3},

   1\2/3

return[3,2,1].

Note: Recursive solution is trivial, could you do it iteratively?

import java.util.*;
public class Solution {public ArrayList<Integer> postorderTraversal(TreeNode root) {ArrayList<Integer> result = new ArrayList<>();if(root == null) {return result;}LinkedList<TreeNode> stack = new LinkedList<TreeNode>();stack.push(root);TreeNode pre = null;while(!stack.isEmpty()) {TreeNode temp = stack.peek();if(temp.left == null && temp.right == null) {result.add(temp.val);stack.pop();pre = temp;} else if((pre != null && (pre == temp.left || pre == temp.right))) {result.add(temp.val);stack.pop();pre = temp;} else {if(temp.right != null) {stack.push(temp.right);}if(temp.left != null) {stack.push(temp.left);}}}return result;}
}

题目描述

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree{1,#,2,3},

   1\2/3

return[1,2,3].

Note: Recursive solution is trivial, could you do it iteratively?

import java.util.*;
public class Solution {public ArrayList<Integer> preorderTraversal(TreeNode root) {ArrayList<Integer> result = new ArrayList<>();if(root == null) {return result;}LinkedList<TreeNode> stack = new LinkedList<TreeNode>();stack.push(root);while(!stack.isEmpty()){TreeNode temp = stack.pop();result.add(temp.val);if(temp.right != null)stack.push(temp.right);if(temp.left != null)stack.push(temp.left);}return result;}
}