sorted cookbook

Given an array where elements is sorted in ascending order, convert it to a height balanced BST.Idea: Convert a sorted array into a highly balanced two-fork search tree. The idea is to use the middle value as the root node, and then the left and right arrays are the left and right sub-trees. Recursive solution.The code is as follows:/** * Definition for a binary tree node. * public class TreeNode {* in

Given a singly linked list where elements is sorted in ascending order, convert it to a height balanced BST.Idea: The difference between this problem and the sorted array is that the list does not know the length and the value above. The overall idea is the middle value as the root node, but requires a little strategy, otherwise it will time out.After the whole length is traversed, the length is saved so th

November 2014-a given integer array is sorted by this integer group to minimize the value of each element in the sorted array. My general idea is to convert this integer array into a string array, and then sort the array for the first time using the compareto method of the string class. The sorting result is exactly sorted by lexicographically, the Lexicographic

http://www.geeksforgeeks.org/generate-all-possible-sorted-arrays-from-alternate-elements-of-two-given-arrays/Given-sorted arrays A and B, generate all possible arrays such the first element is the taken from A then from B then fro M A and so on increasing order till the arrays exhausted. The generated arrays should end with a element from B.For ExampleA = {Ten, 20, 30} B = {1, 5;The resulting arrays is:10 2

translation给定两个排序的整型数组nums1和nums2，将nums2合并到nums1成一个排序数组。批注：你可以假设nums1中有足够的空间（空间大于或等于m+n）来存放来自nums2的额外元素。nums1和nums2的初始空间分别是m和n。Originalgiven, sorted integer Arrays nums1 and nums2, merge nums2 into nums1 Span class= "Hljs-keyword" >as one sorted array. Note:you may assume that nums1 have enough space ( Size that is Greater or equal to m + N) to hold additional elements from nums2. The number o

LeetCode 83 Remove Duplicates from Sorted List (Remove duplicate elements from Sorted List )(*)Translation Given a sorted Linked List, delete all repeated elements so that each element appears only once. For example, if 1-> 1-> 2 is specified, 1-> 2 is returned. Given 1-> 1-> 2-> 3-> 3, 1-> 2-> 3 is returned. Original Given a

Given sorted integer Arrays nums1 and nums2, merge nums2 into nums1 as one sorted Array.Note:Assume that nums1 have enough space (size that's greater or equal to m + n) to hold add Itional elements from nums2. The number of elements initialized in nums1and nums2 is m and n respectively.Idea: This problem obviously merge the sort of shadow, just need to be flexible, from the back to the order of the forward.

Given a sorted linked list, delete all nodes that has duplicate numbers, leaving only distinct numbers from the Original list.For example,Given1->2->3->3->4->4->5, Return1->2->5.Given1->1->1->2->3, Return2->3.The difference between this and the remove duplicate from sorted list is that, in this case, as long as the node is duplicated, the value of the equal node will be deleted, leaving one in the title doe

Suppose a sorted array is rotated on some pivot unknown to you beforehand.(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2 ).You is given a target value to search. If found in the array is return its index, otherwise return-1.Assume no duplicate exists in the array.The equivalent of folding the next part of the array to the front of the array, which is essentially a dichotomy, here is also a rule to follow: first of all to get a turning point, that is,

Follow up for "Remove duplicates":What if duplicates is allowed at the most twice?For example,Given sorted array nums = [1,1,1,2,2,3] ,Your function should return length =5, with the first five elements ofNumsBeing1,1,2,2and3. It doesn ' t matter what are you leave beyond the new length.Idea: The subject allows the array to have a maximum of two repetitions, so you need to add a judgment, whether to two. The specific code is as follows:public class So

Java Object instanceof keyword exercise: judge whether the set of the same person is sorted by age. If the same age is sorted by name in alphabetical order, Comparator, Package com. swift; public class Same_Person_Test {public static void main (String [] args) {/** Object determines whether the Object is the same Person */Person per1 = new Person ("zhangsan", 30 ); person per2 = new Person ("lisi", 27); Per

Given a sorted linked list, delete all duplicates such this each element appear only once.For example,Given1->1->2, return1->2.Given1->1->2->3->3, return1->2->3.Remove duplicate nodes in a single linked list public class Solution { public ListNode DeleteDuplicates(ListNode head) { if (head == null) { return null; } ListNode node = head; ListNode nextNode = null;

Given a sorted array, remove the duplicates in place such, all element appear only once and return the new L Ength.Do the allocate extra space for another array, and you must does this on place with constant memory.For example,Given input Array A =[1,1,2],Your function should return length =2, and A is now[1,2].Test instructions: Remove duplicate elements, if duplicate only one, return length.Idea: Iterate over the array, with different, overlay the s

Given a sorted linked list, delete all duplicates such this each element appear only once.Example 1:Input:1->1->2output:1->2Example 2:Input:1->1->2->3->3output:1->2->3A simple list problem can be written in both recursive and iterative forms. Specific ideas:The first step is to find the node with the first node value and the node value that the current header refers to.In the second step, the next point of the current table head node points to the fou

Title DescriptionGiven a sorted array, you need to Delete the repeating element in place so that each element appears up to two times, returning the new length of the array after it is removed.Instead of using extra array space, you must modify the input array in place and complete it with O (1) Extra space.Example 1:Nums [1,1,1,2,2,3] 5 . 1, 1, 2, 2, You do not need to consider elements that are beyond the new length in the array.Example 2:Nums

17. merge two sorted linked lists, and 17 merge two sorted lists. When the code tries to access the memory pointed to by a null pointer, the program crashes, leading to robustness issues. Therefore, the empty linked list must be processed separately. ListNode * Merge (ListNode * pHead1, ListNode * pHead2){If (pHead1 = NULL)Return pHead2;Else if (pHead2 = NULL)Return pHead1;ListNode * pMergedHead = NUL

There are two sorted arrays nums1 and nums2 of size m and n respectively. find the median of the two sorted arrays. the overall run time complexity shoshould be O (log (m + n ))., respectively Solution: Combine Two arrays and create a Map object to store the ordered key-value pairs. The key is the serial number and the value is the array value. The median result is discussed in two cases: 1. m + n is an odd

The idea first traverses from left to right, then sets a max, and if Array[i] is greater than Max during the Traverse, specify K to record the position if it is less than Max. Then from right to left to traverse, set a min, in the process of traversal array[i] less than min, then the displacement min, if greater than min, then specify J record the location. So the j~~k between the need to sort. As in the example above, Max is 1, then traverse max=1,array[i]=1, displace max,max=1max=1,array[i]=5,

PHP multi-dimensional arrays are sorted by the values of one field, and php multi-dimensional arrays are sorted. In normal times, simple one-dimensional arrays or simple array sorting are not described here. This is mainly used to sort projects based on one of the multi-dimensional arrays.The php function used is: array_multisort. Train of Thought: Get the fields you need to sort as one-dimensional array