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 Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 10⁵) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. 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 Data Next

where?Address?is the position of the node,?Data?is an integer, and?Next?is the position of the next node.

Output Specification:

For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:

00100 6 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218

Sample Output:

00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1 #include<cstdio> #include<algorithm> using namespace std; const int maxn = 100100; struct Node{int address,data,next;int order; }node[maxn];bool cmp(Node a,Node b){return a.order < b.order; } int main(){int i,address;for(i = 0; i < maxn; i++){node[i].order = maxn;}int begin,n,k; //起始節點地址，節點數目，分組數 scanf("%d%d%d",&begin,&n,&k);for(i = 0; i < n; i++){scanf("%d",&address);scanf("%d%d",&node[address].data,&node[address].next);node[address].address = address;}int p = begin,count = 0;while(p != -1){node[p].order = count++;p = node[p].next;}sort(node,node+maxn,cmp);n = count; //因為count=0占一個有效節點，退出循環時，count值就是有效節點 for(i = 0; i < n/k; i++){ //枚舉完整的n/k塊 for(int j = (i+1)*k - 1; j > i*k; j-- ){ //每塊的第i個倒著輸出，剩余最后一個節點 printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address);}printf("%05d %d ",node[i*k].address,node[i*k].data); //每塊的最后一個節點的前兩項數據 if(i < n/k -1) { //如果是非最后一塊節點 printf("%05d\n",node[(i+2)*k-1].address);}else{ //如果是最后一塊節點 if(n % k == 0) printf("-1\n"); //剛好除整 else{ //如果最后一個節點不規則 printf("%05d\n",node[(i+1)*k].address);for(i = n/k*k; i < n; i++){printf("%05d %d ",node[i].address,node[i].data);if(i < n - 1){printf("%05d\n",node[i+1].address);}else{printf("-1\n");}// else} // for(i)}//else}//else} //for(i)return 0; }

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轉載于:https://www.cnblogs.com/wanghao-boke/p/8550233.html

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