(Stacknum==2){ S->Array[--S->Top2]=X; } return 1; } ElementType Top_Pop( Stack S, int Stacknum ){ int p; if(Stacknum==1){ if(IsEmpty(S,1)) return ERROR; p=S->Array[S->Top1--]; } else if(Stacknum==2){ if(IsEmpty(S,2)) return ERROR; p=S->Array[S->Top2...
LeetCode 新手科学刷题顺序:Array 数组(283/27) --> Linked list 链表 (21/203/206)--> 队列 Queue --> 栈 Stack --> ...在链表之前,我们已经学了LeetCode最基础的数据机构数组 Array 以及配套的3…
0025. Reverse Nodes in K Group 0026. Remove Duplicates From Sorted Array 0027. Remove Element 0028. Find the Index of the First Occurrence in a String 0029. Divide Two Integers 0030. Substring With Concatenation of All Words 0031. Next Permutation 0032. Longest Valid Parentheses 0033. Search...
Given a non-empty array of numbers, a0, a1, a2, … , an-1, where 0 ≤ ai < 231. Find the maximum result of ai XOR aj, where 0 ≤ i, j < n. Could you do this in O(n) runtime? Example: Input: [3, 10, 5, 25, 2, 8] Output: 28 Explanation: The maximum result is ...
本题难度: Medium/Easy Topic: Data Structure - stack/queue Description As the title described, you should only use two stacks to implement a queue's actions. The queue should support push(element), pop() and top() where pop is pop the first(a.k.a front) element in the queue. Both...
In the code, we check if m is larger than n to garentee that the we always know the smaller array, for coding simplicy. 中文翻译: 该方法的核心是将原问题转变成一个寻找第k小数的问题(假设两个原序列升序排列),这样中位数实际上是第(m+n)/2小的数。所以只要解决了第k小数的问题,原问题也得以...
LeetCode - 4. Median of Two Sorted Arrays(二分) 题目链接 题目 解析 假设两个数组的中间位置为k,其中k=(n1 + n2 + 1)/2,只要找到中间位置这个值,也就找到了中位数,所以我们可以把问题转换成查找两个数组中第 k 大的数。 如果是总数是偶数,那么如下图,我们的中位数肯定是(Ck-1 + CK) / 2;而...
class Solution { public: ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) { stack<int> s1; stack<int> s2; while(l1!=NULL) { s1.push(l1->val); l1 = l1->next; } while(l2!=NULL) { s2.push(l2->val); l2 = l2->next; } ListNode* list = new ListNode(0); int sum = 0...
按照题意做即可。 代码# Go packageleetcode/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */funcmergeTwoLists(l1*ListNode,l2*ListNode)*ListNode{ifl1==nil{returnl2}ifl2==nil{returnl1}ifl1.Val<l2.Val{l1.Next=mergeTwoLists(l1.Next,l2...
Stack s2= new Stack(); ListNode result=null,t=null; while(l1!=null) { s1.push(l1); l1=l1.next; } while(l2!=null) { s2.push(l2); l2=l2.next; } while(!s1.isEmpty() && !s2.isEmpty()) { if(s1.peek().val > s2.peek().val) ...