题意:给出两条链表的首地址以及若干节点的地址,数据,下一个节点的地址,求两条链表的首个共用节点的地址.如果两条链表没有共用节点,则输出-1.思路步骤1:由于地址的范围很小,因此可以直接用静态链表,但是依照题目的要求,在结点的结构体中再定义一个int型变量flag,表示结点是否在第一条链表中出现,是则为l,不是为 - 1.步骤2:由题目给出的第一条链表的首地址出发遍历第一条链表,将经过的所有结点的flag值赋为1.接下来枚举第二条链表,当出现第一个fag值为1的结点,说明是第一条链表中出现过的结果…

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…

1032 Sharing (25)(25 分) 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 "…

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 a…

其实就是链表求交: #include <iostream> #include <cstdio> #include <cstdlib> #include <unordered_map> using namespace std; class Node { public: Node() : data(), next() {} char data; int next; }; void print_list(unordered_map<int, Node*>…

题目描写叙述 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 s…

题意:给出两条链表的首地址以及若干个节点的的地址.数据.下一个节点的地址,求两条链表的首个共用节点的地址.如果两条链表没有共用节点,则输出-1. 思路:使用静态链表,首先遍历一遍第一个链表并进行标记.然后遍历第二个链表,并检查标记元素,得出结果,进行输出. 代码如下: ```cpp//所用解法不涉及节点的数据及其地址,//故不需要在节点中来存储 #include<iostream>#include<algorithm>using namespace std;const int ma…

题目 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 Figu…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…

1032 Sharing (25)(25 分) 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 "…