随笔列表第4页 - PJCK

2019年4月28日

(双指针二分) leetcode 167. Two Sum II - Input array is sorted

摘要： Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number. The function 阅读全文

posted @ 2019-04-28 18:26 PJCK 阅读(132) 评论(0) 推荐(0) 编辑

(双指针) leetcode 485. Max Consecutive Ones

摘要： Given a binary array, find the maximum number of consecutive 1s in this array. Example 1: Note: The input array will only contain 0 and 1. The length 阅读全文

posted @ 2019-04-28 17:35 PJCK 阅读(148) 评论(0) 推荐(0) 编辑

(双指针) leetcode 27. Remove Element

摘要： Given an array nums and a value val, remove all instances of that value in-place and return the new length. Do not allocate extra space for another ar 阅读全文

posted @ 2019-04-28 17:15 PJCK 阅读(167) 评论(0) 推荐(0) 编辑

2019年4月27日

(String) leetcode 67. Add Binary

摘要： Given two binary strings, return their sum (also a binary string). The input strings are both non-empty and contains only characters 1or 0. Example 1: 阅读全文

posted @ 2019-04-27 23:13 PJCK 阅读(143) 评论(0) 推荐(0) 编辑

2019年4月26日

(数组) leetcode 66. Plus One

摘要： Given a non-empty array of digits representing a non-negative integer, plus one to the integer. The digits are stored such that the most significant d 阅读全文

posted @ 2019-04-26 21:39 PJCK 阅读(120) 评论(0) 推荐(0) 编辑

(N叉树 BFS) leetcode429. N-ary Tree Level Order Traversal

摘要： Given an n-ary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). For example, given a 3-ary tree: 阅读全文

posted @ 2019-04-26 19:17 PJCK 阅读(143) 评论(0) 推荐(0) 编辑

(N叉树递归) leetcode 590. N-ary Tree Postorder Traversal

摘要： Given an n-ary tree, return the postorder traversal of its nodes' values. For example, given a 3-ary tree: Return its postorder traversal as: [5,6,3,2 阅读全文

posted @ 2019-04-26 19:15 PJCK 阅读(227) 评论(0) 推荐(0) 编辑

(N叉树递归) leetcode589. N-ary Tree Preorder Traversal

摘要： Given an n-ary tree, return the preorder traversal of its nodes' values. For example, given a 3-ary tree: Return its preorder traversal as: [1,3,5,6,2 阅读全文

posted @ 2019-04-26 19:12 PJCK 阅读(146) 评论(0) 推荐(0) 编辑

(N叉树 DFS 递归 BFS) leetcode 559. Maximum Depth of N-ary Tree

摘要： Given a n-ary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest le 阅读全文

posted @ 2019-04-26 18:44 PJCK 阅读(102) 评论(0) 推荐(0) 编辑

(BST 递归) leetcode98. Validate Binary Search Tree

摘要： Given a binary tree, determine if it is a valid binary search tree (BST). Assume a BST is defined as follows: The left subtree of a node contains only 阅读全文

posted @ 2019-04-26 16:07 PJCK 阅读(183) 评论(0) 推荐(0) 编辑

(二叉树递归) leetcode 145. Binary Tree Postorder Traversal

摘要： Given a binary tree, return the postorder traversal of its nodes' values. Example: Follow up: Recursive solution is trivial, could you do it iterative 阅读全文

posted @ 2019-04-26 13:18 PJCK 阅读(87) 评论(0) 推荐(0) 编辑

(二叉树递归) leetcode94. Binary Tree Inorder Traversal

摘要： Given a binary tree, return the inorder traversal of its nodes' values. Example: Follow up: Recursive solution is trivial, could you do it iteratively 阅读全文

posted @ 2019-04-26 13:13 PJCK 阅读(81) 评论(0) 推荐(0) 编辑

(二叉树递归) leetcode 889. Construct Binary Tree from Preorder and Postorder Traversal

摘要： Return any binary tree that matches the given preorder and postorder traversals. Values in the traversals pre and post are distinct positive integers. 阅读全文

posted @ 2019-04-26 10:35 PJCK 阅读(175) 评论(0) 推荐(0) 编辑

(二叉树递归) leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal

摘要： Given preorder and inorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, 阅读全文

posted @ 2019-04-26 09:12 PJCK 阅读(139) 评论(0) 推荐(0) 编辑

(二叉树递归) leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal

摘要： Given inorder and postorder traversal of a tree, construct the binary tree. Note:You may assume that duplicates do not exist in the tree. For example, 阅读全文

posted @ 2019-04-26 09:09 PJCK 阅读(180) 评论(0) 推荐(0) 编辑