1074 Reversing Linked List (25)(25 分) Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you mus…

1074 Reversing Linked List (25 分)   Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must outp…

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L.  For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6. Input Specifica…

题目 Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6. Input Specifi…

模拟题,注意当k == 1 与 k == n时情况 #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <map> using namespace std; const int N = 100005; struct Node { int pre; int value; int lat; }node[N]; int orde…

简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; +; struct Node { ]; ]; ]…

题意说的很清楚了,这种题的话,做的时候最好就是在纸上自己亲手模拟一下,清楚一下各个指针的情况, 这样写的时候就很清楚各个指针变量保存的是什么值. PS:一次AC哈哈,所以说自己动手在纸上画画还是很有好处的~ #include <iostream> #include <cstdio> #include <algorithm> #include <string.h> #include <cmath> #include <queue> us…

#include <cstdio> #include <cstdlib> #include <iostream> #include <unordered_map> using namespace std; class Node { public: int data; int next; Node() : data(), next() {} Node(int d, int n):data(d), next(n){} }; int reverse_help(in…

题意: 每k个元素反转链表,不足k个就不反转.如原链表为1→2→3→4→5→6,k=3,则反转后的链表为3→2→1→6→5→4:若k=4,则反转后的链表为4→3→2→1→5→6. 思路: 这题会比较烦,写代码前一定要现在纸上理清思路,写出关键代码,不然出了错再改来改去真的很浪费时间,要是考试的话估计心态就蹦了.本题是比较经典的“反转链表(指反转整条链表)”的升级版,但做法是一样的.我们可以把“反转”这一动作单独抽象成一个函数,然后遍历链表,每遍历k个结点(假设是pi...pj),就把pi...p…

题意: 输入链表头结点的地址(五位的字符串)和两个正整数N和K(N<=100000,K<=N),接着输入N行数据,每行包括结点的地址,结点的数据和下一个结点的地址.输出每K个结点局部反转的链表. trick: 测试点6包含一些不在起点这条链表上的结点. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include<bits/stdc++.h> using namespace std; map<string,pair<i…