[PAT解题报告] 反转链表 (25)
转载from http://tech-wonderland.net/blog/pat-1074-reversing-linked-list.html
其实就是个模拟题, 改怎么反转就怎么反转, 就是测试用例比较恶心, 注意输入可能不止一个链表的情况, 这种情况下, 要忽略那些不在该链表（输入中的head对应的那个链表）上的节点, 也就是有效节点数（输出的节点个数）比输入中的N要小. 题目的输入使用了数组形式的链表, 可以事先就开好一个足够大的数组, 反转的过程就是一般的反转了, 注意记录下反转以后新的链表头尾, 还有不要忘记把每段反转以后的新的链表接起来. AC代码如下:
#include <cstdio>
struct Node {
int iData;
int iNext;
};
const int iMaxLen = 100004;
int iHead;
int N, K;
Node NodeList[iMaxLen];
void doPrint(int iListHead) {
while(-1 != iListHead) {
if(NodeList[iListHead].iNext == -1) {
printf("%.5d %d -1\n",
iListHead,
NodeList[iListHead].iData);
break;
}
printf("%.5d %d %.5d\n",
iListHead,
NodeList[iListHead].iData,
NodeList[iListHead].iNext);
iListHead = NodeList[iListHead].iNext;
}
}
int reverseK(int iStartHead, int iK,
int *pHead, int *pTail) {
if(-1 == iStartHead || iK <= 1)
return -1;
if(2 == iK) {
int node1 = iStartHead;
int node2 = NodeList[iStartHead].iNext;
NodeList[node1].iNext = NodeList[node2].iNext;
NodeList[node2].iNext = node1;
*pHead = node2;
*pTail = node1;
return 0;
}
*pTail = iStartHead;
int node1 = iStartHead;
int node2 = NodeList[iStartHead].iNext;
int nodeTmp = -1;
for(int i = 0; i < iK - 1; ++i) {
nodeTmp = NodeList[node2].iNext;
NodeList[node2].iNext = node1;
node1 = node2;
node2 = nodeTmp;
}
*pHead = node1;
NodeList[*pTail].iNext = node2;
return 0;
}
void gao() {
int iNodeTmp = iHead;
int iEffectiveLength = 1;
while(-1 != NodeList[iNodeTmp].iNext) {
++iEffectiveLength;
iNodeTmp = NodeList[iNodeTmp].iNext;
}
if(K > iEffectiveLength) {
doPrint(iHead);
}
N = iEffectiveLength;
int iNewHead;
if(K > 1) {
int iTheHead, iTheTail;
int iLastTail;
// first init reverse to decide the new head
reverseK(iHead, K, &iTheHead, &iTheTail);
iNewHead = iTheHead;
iLastTail = iTheTail;
int iReverseCount = N / K - 1;
for(int i = 0; i < iReverseCount; ++i) {
reverseK(NodeList[iTheTail].iNext, K, &iTheHead, &iTheTail);
NodeList[iLastTail].iNext = iTheHead;
iLastTail = iTheTail;
}
}
else
iNewHead = iHead;
doPrint(iNewHead);
}
int main() {
for(int i = 0; i < iMaxLen; ++i) {
NodeList[i].iData = 0;
NodeList[i].iNext = -1;
}
scanf("%d %d %d", &iHead, &N, &K);
int iSingleNodeAddress, iSingleNodeData, iSingleNodeNext;
for(int i = 0; i < N; ++i) {
scanf("%d %d %d", &iSingleNodeAddress, &iSingleNodeData, &iSingleNodeNext);
NodeList[iSingleNodeAddress].iData = iSingleNodeData;
NodeList[iSingleNodeAddress].iNext = iSingleNodeNext;
}
gao();
return 0;
}
(1) 这次PAT最坑的一个题目吧, 输入可能不止一个链表, 有些节点不在要反转的那个链表上, 输出的时候应当忽略. 比如00100 6 4(2) 另外注意指针的使用, 还有特例K = 1, K = N之类的情况, 因为上面的特殊坑点, 有可能是打断了的多个链表, 我们的K也有可能大于有效节点数.
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 -1
99999 5 68237
12309 2 33218
