有序数组的归并 链表的合并lianbiao

有序数组合并

有两个递增数组，第一个数组的大小为 m+n   有m个有效数，第二个数组大小为n，有效数为n.试将两个数组合并到第一个数组，且保持数组一的有序。

看到这个题特别容易想到链表的合并，但是若按照链表的正序方法，时间复杂度为O(m*n),效率太慢

所以应该从后向前插入。时间复杂度为O(m+n)

如下代码，含注释。

#include<iostream>#include<assert.h>using namespace std;#define m 4#define n 5void ComOrdArray(int* a1, int* a2, int n1,int n2){assert(a1||a2);//判断是否有效size_t i = n1 + n2-1;  //被放值位置size_t j = 0;          //n1-j为比较位置while (n2&&(j<n1))  //注意j<n1  当数组一中的数值都被移到后面时，就可跳出，直接进行对应赋值 //防止if (*(a1 + n1 - 1-j) <= *(a2 + n2 - 1))时出错{if (*(a1 + n1 - 1-j) <= *(a2 + n2 - 1))//当被插入数大于或等于时，直接插入倒数对应位{*(a1 + i) = *(a2 + n2 - 1);--i;}else                             //else   将数组一的数向后移动{while (*(a1 + n1 - 1 - j) > *(a2 + n2 - 1)){*(a1 + i) = *(a1 + n1 - 1 - j);--i;++j;}*(a1 + i) = *(a2 + n2 - 1);   //找到插入位置  插入--i;}      n2--;}for (size_t num = 0; num < n2;num++)    {*(a1 + num) = *(a2 + num);}}int main(){int a[m + n] = {5,6,7,15};int aa[n] = {2,3,6,8,10};ComOrdArray(a,aa,m,n);for (size_t i = 0; i < m + n; ++i){cout << a[i] << " ";}return 0;}

有序链表合并

两个有序链表的合并在合并时需要注意的不多，但是对于链表的析构必须要处理好。将s2插入到s1中，由于在合并时并没有创建新的节点，所以原先s2的结构会有改变，且s1含有所有的s2节点，所以将s2也指向s1，两个链表实际管理的是同一个，所以在析构时要特别注意。

实现代码：

void _ComSortList( List<T> &s2){if (_headlist == NULL){_headlist = s2._headlist;return;}Node* cur1 = _headlist;Node* cur2 =s2._headlist;Node* next2 = cur2;Node* prev1 = NULL;while (cur2&&cur1){if (cur2->_data > cur1->_data){prev1 = cur1;cur1 = cur1->_next;}else{next2 = cur2->_next;if (prev1 == NULL){_headlist = cur2;prev1 = cur2;}elseprev1->_next = cur2;cur2->_next = cur1;cur2 = next2;}}if (cur1 == NULL){prev1->_next = cur2;}s2._headlist = _headlist;_count++;s2._count--;}

具体代码：

#include<iostream>#include<stack>using namespace std;template<class T>class ListNode{public:ListNode(const T &data):_data(data), _next(NULL){}T _data;ListNode* _next;};template<class T>class List{public:typedef ListNode<T> Node;List():_headlist(NULL),_count(1){}void PushBack(const T& data){Node* tmp = new Node(data);if (_headlist == NULL){_headlist = tmp;}else{Node* cur = _headlist;while (cur->_next){ cur = cur->_next;}cur->_next = tmp;}}~List(){if (_headlist&&_count){delete _headlist;}_headlist = NULL;_count--;}void Printf(){if (_headlist == NULL)return;Node* cur = _headlist;while (cur){cout << cur->_data << " ";cur = cur->_next;}cout << endl;}void RevPrintfList1()//利用stack{if (_headlist == NULL)return;Node* cur = _headlist;stack<T> st;while (cur){st.push(cur->_data);cur = cur->_next;}cout << "the relist is :" << endl;while (!st.empty()){cout << st.top() << " ";st.pop();}cout << endl;}void RevPrintfList2()//递归{if (_headlist == NULL)return;cout << "the relist is :" << endl;_RevPrintfList2(_headlist);cout << endl;}void _ComSortList( List<T> &s2){if (_headlist == NULL){_headlist = s2._headlist;return;}Node* cur1 = _headlist;Node* cur2 =s2._headlist;Node* next2 = cur2;Node* prev1 = NULL;while (cur2&&cur1){if (cur2->_data > cur1->_data){prev1 = cur1;cur1 = cur1->_next;}else{next2 = cur2->_next;if (prev1 == NULL){_headlist = cur2;prev1 = cur2;}elseprev1->_next = cur2;cur2->_next = cur1;cur2 = next2;}}if (cur1 == NULL){prev1->_next = cur2;}s2._headlist = _headlist;_count++;s2._count--;}private:void _RevPrintfList2(Node* cur){if (cur == NULL)return;_RevPrintfList2(cur->_next);cout << cur->_data<<" ";}Node* _headlist;int _count;};#define n1 4#define n2 5void ComSortList(){int a1[n1] = { 5, 6, 7, 15 };int a2[n2] = { 2, 3, 6, 8, 10 };List<int> s1;List<int> s2;for (size_t i = 0; i < n1; ++i){s1.PushBack(a1[i]);}s1.Printf();for (size_t i = 0; i < n2; ++i){s2.PushBack(a2[i]);}s2.Printf(); cout << endl;s1._ComSortList(s2);s1.Printf();}