#include <iostream> using namespace std; struct ListNode { int m_nValue; ListNode *m_pNext; }; ListNode* CreateList(int val) { ListNode *pHead = new ListNode; pHead->m_nValue = val; pHead->m_pNext = NULL; return pHead; } void InsertNode(ListNode **pHead, int val) { ListNode *pNode = new ListNode; pNode->m_nValue = val; pNode->m_pNext = NULL; while ((*pHead)->m_pNext != NULL) { (*pHead) = (*pHead)->m_pNext; } (*pHead)->m_pNext = pNode; (*pHead) = pNode; } void PrintList(ListNode *pHead) { while (pHead != NULL) { cout<<pHead->m_nValue<<" "; pHead = pHead->m_pNext; } cout<<endl; } ListNode* Reverse(ListNode *pHead) { if (pHead == NULL || pHead->m_pNext == NULL) { return pHead; } ListNode *pPre = NULL; ListNode *pCurrent = pHead; ListNode *pPost = pHead->m_pNext; while (pCurrent->m_pNext != NULL) { pCurrent->m_pNext = pPre; pPre = pCurrent; pCurrent = pPost; pPost = pPost->m_pNext; } pCurrent->m_pNext = pPre; return pCurrent; } ListNode* ReverseList(ListNode *pHead, int k) { if (pHead==NULL || pHead->m_pNext==NULL) { return pHead; } ListNode *pPre = NULL; ListNode *pCurrent = pHead; ListNode *pPost = pHead->m_pNext; ListNode *pStart = NULL; ListNode *pEnd = NULL; int n = 0; pEnd = pCurrent; pEnd->m_pNext = NULL; while (pPost != NULL) { ++n; if (n == (k+1)) { pStart = pPre; pEnd->m_pNext = ReverseList(pCurrent, k); return pStart; } else { pCurrent->m_pNext = pPre; pPre = pCurrent; pCurrent = pPost; pPost = pPost->m_pNext; } } pCurrent->m_pNext = pPre; pStart = Reverse(pCurrent); return pStart; } int main() { ListNode *pHead = NULL; ListNode *head = NULL; int n; cout<<"輸入鏈表中節點的個數 n:"<<endl; cin>>n; cout<<"請輸入n個整數值:"<<endl; for (int i=0; i<n; ++i) { int data; cin>>data; if (pHead == NULL) { pHead = CreateList(data); head = pHead; } else { InsertNode(&pHead, data); } } int k; cout<<"請輸入k:"<<endl; cin>>k; head = ReverseList(head, k); PrintList(head); system("pause"); return 0; }
2.
#define false 0 #define success 1 int getcurrentms() { struct timeval tv; gettimeofday(&tv,NULL); return tv.tv_sec*1000+tv.tv_usec/1000; //得到毫秒數 } bool count_access() { static int count=0; static int time_ms_old=0,time_ms_now; if(count==0) { time_ms_old=getcurrentms(); } count++; access(); if(count>=R) { time_ms_now=getcurrentms(); if(time_ms_now-time_ms_pld>=1000) return false; else return success; } return success; }
3.
解: 思路:從矩陣的右上方開始判斷即可,每次可以消除一行或一列,詳見劍指offer一書.
4.
#include <iostream> #include <stack> using namespace std; template <class T> class Queue { public: Queue() { } ~Queue() { } void add(const T& t); T remove(); private: stack<T> s1; stack<T> s2; }; template <class T> void Queue<T>::add(const T& t) { s1.push(t); } template <class T> T Queue<T>::remove() { if (s2.size() <= 0) { while (s1.size() > 0) { T t = s1.top(); s2.push(t); s1.pop(); } } if (s2.size() == 0) { throw new exception("empty queue"); } T t = s2.top(); s2.pop(); return t; } int main() { Queue<char> q; q.add('A'); q.add('B'); q.add('C'); cout<<q.remove()<<endl; cout<<q.remove()<<endl; cout<<q.remove()<<endl; system("pause"); return 0; }
public class MaxConString { /** * 計算兩字串最大公用字串長度 */ public static void main(String[] args) { char[] s1 = "jiajiangayaoyao".toCharArray(); //測試資料 char[] s2 = "jiangyaoyao".toCharArray(); int c = new MaxConString().getCount(s1, s2); System.out.println("兩字串的共同字串長度為:"+c); } private int getSubCount(char[] s1,char[] s2, int i ,int j){//計算兩字串從s1的第i位置s2的第j位置的之後字串長度 //如“abc”和“ab”則返回conut為2 int count=1; while(++i<s1.length&&++j<s2.length&&s1[i]==s2[j]){ count++; } return count; } private int getCount(char[]s1,char[]s2){ //計算兩字串的共同字串長度 int count = 0; for(int i=0;i<s1.length;i++) for(int j=0;j<s2.length;j++) if(s1[i]==s2[j]){ if(this.getSubCount(s1, s2, i, j)>count) count = this.getSubCount(s1, s2, i, j); } return count; } }