給定一個(gè)常數(shù) K 以及一個(gè)單鏈表 L，請(qǐng)編寫(xiě)程序?qū)?L 中每 K 個(gè)結(jié)點(diǎn)反轉(zhuǎn)。例如：給定 L 為 1→2→3→4→5→6，K 為 3，則輸出應(yīng)該為 3→2→1→6→5→4；如果 K 為 4，則輸出應(yīng)該為 4→3→2→1→5→6，即最后不到 K 個(gè)元素不反轉(zhuǎn)。
輸入格式：
每個(gè)輸入包含 1 個(gè)測(cè)試用例。每個(gè)測(cè)試用例第 1 行給出第 1 個(gè)結(jié)點(diǎn)的地址、結(jié)點(diǎn)總個(gè)數(shù)正整數(shù) N (≤105)、以及正整數(shù) K (≤N)，即要求反轉(zhuǎn)的子鏈結(jié)點(diǎn)的個(gè)數(shù)。結(jié)點(diǎn)的地址是 5 位非負(fù)整數(shù)，NULL 地址用 ?1 表示。
To store English words, one method is to use linked lists and store a word letter by letter. To save some space, we may let the words share the same sublist if they share the same suffix. For example, loading and being are stored as showed in Figure 1.
Figure 1
You are supposed to find the starting position of the common suffix (e.g. the position of i in Figure 1).
Input Specification:
Each input file contains one test case. For each case, the first line contains two addresses of nodes and a positive N (≤105), where the two addresses are the addresses of the first nodes of the two words, and N is the total number of nodes. The address of a node is a 5-digit positive integer, and NULL is represented by ?1.
Then N lines follow, each describes a node in the format:
Address Data Next
whereAddress is the position of the node, Data is the letter contained by this node which is an English letter chosen from { a-z, A-Z }, and Next is the position of the next node.
Output Specification:
For each case, simply output the 5-digit starting position of the common suffix. If the two words have no common suffix, output -1 instead.
Sample Input 1:
11111 22222 9
67890 i 00002
00010 a 12345
00003 g -1
12345 D 67890
00002 n 00003
22222 B 23456
11111 L 00001
23456 e 67890
00001 o 00010
Sample Output 1:
67890
Sample Input 2:
00001 00002 4
00001 a 10001
10001 s -1
00002 a 10002
10002 t -1
Sample Output 2:
-1
我的
#include <iostream>
#include <map>
using namespace std;
int main(void)
{
int start1 = 0, start2 = 0, n = 0;
scanf("%d %d %d", &start1, &start2, &n);
map<int, int> normal;
for (int i = 0;i < n; i) {
int address = 0, c = 0, next = 0;
scanf("%d %c %d", &address, &c, &next);
normal[address] = next;
}
map<int, bool> link1;
int now = start1;
for (;now != -1;now = normal[now]) link1[now] = true;
now = start2;
for (;now != -1;now = normal[now]) if (link1[now]) break;
if (now != -1) printf("d", now);
else printf("-1");
return 0;
}
A linked list consists of a series of structures, which are not necessarily adjacent in memory. We assume that each structure contains an integer key and a Next pointer to the next structure. Now given a linked list, you are supposed to sort the structures according to their key values in increasing order.
Input Specification:
Each input file contains one test case. For each case, the first line contains a positive N (<105) and an address of the head node, where N is the total number of nodes in memory and the address of a node is a 5-digit positive integer. NULL is represented by ?1.
Then N lines follow, each describes a node in the format:
Address Key Next
where Address is the address of the node in memory, Key is an integer in [?105,105], and Next is the address of the next node. It is guaranteed that all the keys are distinct and there is no cycle in the linked list starting from the head node.
Output Specification:
For each test case, the output format is the same as that of the input, where N is the total number of nodes in the list and all the nodes must be sorted order.
#include <iostream>
#include <map>
using namespace std;
int main(void)
{
int n = 0, start = 0;
cin >> n >> start;
map<int, int> normal, addresses, keys;
for (int i = 0;i < n; i) {
int address = 0, key = 0, next = 0;
cin >> address >> key >> next;
normal[address] = next;
addresses[address] = key;
}
int now = start;
for (;now != -1;now = normal[now]) keys[addresses[now]] = now;
printf("%d ", keys.size());
map<int, int>::iterator iter = keys.begin();
for (;iter != keys.end(); iter) printf("d\nd %d ", iter->second, iter->second, iter->first);
printf("-1");
return 0;
}
注意無(wú)效結(jié)點(diǎn)的處理：測(cè)試點(diǎn)1、4
《算法筆記》P279
A1097 Deduplication on a Linked List
Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (≤105) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by ?1.
Then N lines follow, each describes a node in the format:
Address Key Next
where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.
Output Specification:
For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.
