1133. Splitting A Linked List (25)[链表处理]

1. 原题： https://www.patest.cn/contests/pat-a-practise/1133

2. 思路：

题意：链表处理题
将一杂乱的链表按照一定格式（负的在前，大于给定值的在后）重新排列。
思路：
先把每个结点存储在数组里，
然后遍历3次，存入vector里，最后输出即可。

已AC

3. 源码

#include <iostream>#include <vector>using namespace std;struct Node{int local;int data;int next;};Node dt[100000];int main(){//freopen("in.txt", "r", stdin);int N, head, K, tmp;vector<Node> re;scanf("%d %d %d", &head, &N, &K);for (int i = 0; i < N; i++){scanf("%d", &tmp);dt[tmp].local = tmp;scanf("%d %d", &dt[tmp].data, &dt[tmp].next);}int adr = head;while (1)//***找出小于0的{if (adr == -1)break;if (dt[adr].data < 0)re.push_back(dt[adr]);adr = dt[adr].next;}adr = head;while (1){if (adr == -1)break;if (dt[adr].data > -1 && dt[adr].data < (K+1))re.push_back(dt[adr]);adr = dt[adr].next;}adr = head;while (1)//***找出大于K的{if (adr == -1)break;if (dt[adr].data > K)re.push_back(dt[adr]);adr = dt[adr].next;}int size = re.size();for (int i = 0; i < size-1; i++){printf("%05d %d %05d\n", re[i].local, re[i].data, re[i+1].local);}printf("%05d %d -1\n", re[size-1].local, re[size-1].data);return 0;}