l2 gartner

The following is the time to take time to brush the first 5 questions, are the solution of their own, perhaps not the optimal solution, just finishing, convenient for future optimization and promotion1. The Sum of the sums:class Solution: # @return A tuple, (index1, index2) def twosum (self, num, target): = {} for in xrange (len (num)): if dict.get ( Target-num[i], none) = = None := I else: return (Dict[target-num[i]], i)

Problem descriptionMerge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.AlgorithmCode One1 PublicListNode mergetwolists (listnode l1,listnode L2) {2 if(l1==NULL)3 returnL2;4 if(l2==NULL)5 returnL1;6 ListNode l3,p;7 if(l1.vall2.val) {8l3=L1;9p=L1;Tenl1=L1.next; One}Else{ Al3

The first part of the problem is the operation of a list of cases, this article mainly on a number of linked list of some problems(1) Merging two lists that have been sorted //l1, L2 Two pointers constantly moving back ListNode*mergesortedlists (ListNode *l1, ListNode *L2) {ListNode Newhead (0), *pos = Newhead; while(L1 | |L2) { if(!L1)return(Pos->nex

I have written two versions for your reference: Recursive version ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) { return addTwoNumbers(l1, l2, 0); } ListNode *addTwoNumbers(ListNode *l1, ListNode *l2, int carry) { if(l1 == NULL) return addTwoNumbers(l2, carry); if(

, can specify delimiter, output [' zcc ', ' Zhangsan ', ' Lisi ', ' Wangwu ']Print (". Join (S5)) #列表转换成字符串, output zcc Zhangsan Lisi WangwuPrint (len (s)) #测量长度, output 8Print (S.count (' C ')) #计数c出现的次数, Output 23.listL1 = [3,4], 5,6] L2 = [' Zhangsan ', ' John Doe ', ' cc '] l3 = [1,4,7,8,9,0,3]########### #查找值Print (L1[2],type (l1[2))) #输出 (3, 4) Print (L1[0:3],type (l1[0:3)) #输出 [1, 2, (3, 4)] For I in L1:Print (i) #for循环取值########## #list增加值

as the data inside. Print () will get the type that belongs to the element, not the progenitor.tu1= (1)Print (TU1) Gets the elementtu1= (1,)Tu2= (' Alex ',)Tu3= ("the")lists, tuples, strings convert each otherLists and tuples converted to strings must depend on the Join functionLists: Lists and lists can be addedL1= "the"L2= "' Alex ', ' Wusir '"L3=l1+l3Print ()When you loop a list, if you delete some of the elements in the list,Then all elements beh

Original title address: https://oj.leetcode.com/problems/add-two-numbers/Test instructionsYou are given, linked lists representing, and non-negative numbers. The digits is stored in reverse order and all of their nodes contain a single digit. ADD the numbers and return it as a linked list.Input: (2, 4, 3) + (5, 6, 4)Output: 7, 0, 8The idea of solving problems: the basic operation of the linked list. The methods is provided. Method 1 is much slimmer#Definition for singly-linked list.#class ListNo

# Include The line_in_line () function can also be written as follows: Int line_in_line (line L1, line l2) // intersection of line segments (including intersections) {// return (max (l1.u. x, l1.v. x)> = min (l2.u. x, l2.v. x) MAX (l2.u. x, l2.v. x)> = min (l1.u. x, l1.v.

*/Void out_list (sqlist L);/* output set */Void complement_list (sqlist * L1, sqlist * l2);/* Find a complementary set */Void intersect_list (sqlist * L1, sqlist * L2, sqlist * l3);/* intersection calculation */Void union_list (sqlist * L1, sqlist * L2, sqlist * l3);/* calculate the Union set */ /* Main function */Main (){Int choice;/* record the operation selec

, linesegment line){Return (FABS (multiply (line. pt1, line. pt2, point) (Point. x-line.pt1.x) * (point. x-line.pt2.x) (Point. y-line.pt1.y) * (point. y-line.pt2.y) }// Determine the intersection of line segmentsBool intersect (linesegment L1, linesegment l2){Return (max (l1.pt1. X, l1.pt2. X)> = min (l2.pt1. X, l2.pt2. X ))(Max (

The built-in set is easier to remember.L1 = ['B', 'C', 'D', 'B', 'C', 'A', 'a']L2 = List (SET (L1 ))Print L2 There is also a speed difference that is said to be faster and never tested.L1 = ['B', 'C', 'D', 'B', 'C', 'A', 'a']L2 = {}. fromkeys (L1). Keys ()Print L2 Both of them have a disadvantage. Sorting changes aft

to determine whether the list is empty. 2.8 Pop_back and Pop_front (): through Delete the last element, delete the first element by Pop_front (), the sequence must not be empty, and if the list is empty, calling Pop_back () and Pop_front () will cause the program to collapse. 2.9 Assign (): similar to the operation in the vector, there are two cases, the first is: L1.assign (n,val) changes the L1 element to n T (Val). The second case is: L1.assign (L2