15道使用频率极高的基础算法题

1、合并排序，将两个已经排序的数组合并成一个数组，其中一个数组能容下两个数组的所有元素;

合并排序一般的思路都是创建一个更大数组C，刚好容纳两个数组的元素，先是一个while循环比较，将其中一个数组A比较完成，将另一个数组B中所有的小于前一个数组A的数及A中所有的数按顺序存入C中，再将A中剩下的数存入C中，但这里是已经有一个数组能存下两个数组的全部元素，就不用在创建数组了，但只能从后往前面存，从前往后存，要移动元素很麻烦。

void MergeArray(int a[], int alen, int b[], int blen)
{
int len = alen + blen - 1;
alen--;
blen--;
while (alen >= 0 && blen >= 0)
{
if (a[alen] >= b[blen])
{
a[len--] = a[alen--];
}
else
{
a[len--] = b[blen--];
}
}

while (alen >= 0)
{
a[len--] = a[alen--];
}

while (blen >= 0)
{
a[len--] = b[blen--];
}
}

int main()
{
int a[10] = {1,3,5,7,9};
int b[5] = {1,4,6,8,10};
MergeArray(a, 5, b, 5);
for (int i = 0; i < sizeof(a) / sizeof(a[0]); i++)
{
cout<<a[i]<<" ";
}
}

2、合并两个单链表;

struct ListNode{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

ListNode* MergeList(ListNode *head1, ListNode *head2)
{
if (head1 == NULL)
{
return head2;
}
if (head2 == NULL)
{
return head1;
}

ListNode *head; //新链表头结点
if (head1->val <= head2->val)
{
head = head1;
head1 = head1->next;
}
else
{
head = head2;
head2 = head2->next;
}

ListNode *pre = head;
while (head1 != NULL && head2 != NULL)
{
if (head1->val <= head2->val)
{
pre->next = head1;
head1 = head1->next;
}
else
{
pre->next = head2;
head2 = head2->next;
}
pre = pre->next;
}

if (head1 != NULL)
{
pre->next = head1;
}

if (head2 != NULL)
{
pre->next = head2;
}

return head;
}

int main()
{
ListNode *head1 = new ListNode(0);
ListNode *head2 = new ListNode(1);
ListNode *cur1 = head1;
ListNode *cur2 = head2;
for (int i = 2; i < 10; i++)
{
ListNode *newnode = new ListNode(i);
if (i & 1)
{
cur2->next = newnode;
cur2 = newnode;
}
else
{
cur1->next = newnode;
cur1 = newnode;
}
}

ListNode *head = MergeList(head1, head2);
ListNode *temp = head;
while (temp != NULL)
{
cout<<temp->val<<" ";
temp = temp->next;
}
}

3、倒序打印一个单链表;

①递归实现，先递归在打印就变成倒序打印了，如果先打印在调用自己就是顺序打印了。

②借助栈来实现。

struct ListNode{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

void reversePrintNode(ListNode *head)
{
if (head != NULL)
{
reversePrintNode(head->next);
cout<<head->val<<" ";
}
}

void reversePrintNode2(ListNode *head)
{
stack<ListNode*> nodes;
while (head != NULL)
{
nodes.push(head);
head = head->next;
}

while (!nodes.empty())
{
cout<<nodes.top()->val<<" ";
nodes.pop();
}
}

int main()
{
ListNode *head = new ListNode(0);
ListNode *cur = head;
for (int i = 1; i < 10; i++)
{
ListNode *newnode = new ListNode(i);
cur->next = newnode;
cur = cur->next;
}
reversePrintNode2(head);
}

4、给定一个单链表的头指针和一个指定节点的指针，在O(1)时间删除该节点;

#include <iostream>
using namespace std;

struct ListNode{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

void printList(ListNode *head)
{
while (head != NULL)
{
cout<<head->val<<" ";
head = head->next;
}
cout<<endl;
}

void dropList(ListNode *head)
{
if (head == NULL) return;
ListNode *temp;
while (head != NULL)
{
temp = head->next;
delete head;
head = temp;
}
}

ListNode* deleteNode(ListNode *head, ListNode *del)
{
if (head == NULL || del == NULL)
{
return head;
}

if (head == del)
{
ListNode *temp = head->next;
delete head;
head = temp;
}
else if (del->next != NULL)
{
ListNode *next = del->next;
del->val = next->val;
del->next = next->next;
delete next;
}
else
{
ListNode *cur = head;
while (cur != NULL && cur->next != del)
{
cur = cur->next;
}

if (cur != NULL)
{
delete del;
cur->next = NULL;
}
}

return head;
}

int main()
{
for (int i = 0; i < 10; i++)
{
ListNode **nodes = new ListNode*[10];
nodes[0] = new ListNode(0);
ListNode *cur = nodes[0];
for (int j = 1; j < 10; j++)
{
nodes[j] = new ListNode(j);
nodes[j-1]->next = nodes[j];
}

ListNode *head = nodes[0];
ListNode *newhead = deleteNode(head, nodes[i]);
cout<<"删除节点"<<i<<"后的结果: ";
printList(newhead);
dropList(newhead);
delete [] nodes;
}
}

5、找到链表倒数第K个节点;

通过两个指针，两个指针都指向链表的开始，一个指针先向前走K个节点，然后再以前向前走，当先走的那个节点到达末尾时，另一个节点就刚好与末尾节点相差K个节点。

#include <iostream>
using namespace std;

struct ListNode{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};

void printList(ListNode *head)
{
while (head != NULL)
{
cout<<head->val<<" ";
head = head->next;
}
cout<<endl;
}

void dropList(ListNode *head)
{
if (head == NULL) return;
ListNode *temp;
while (head != NULL)
{
temp = head->next;
delete head;
head = temp;
}
}

ListNode* findKthToTail(ListNode *head, int k)
{
if (head == NULL || k == 0)
{
return NULL;
}

ListNode *temp = head;
for (int i = 0; i < k; i++)
{
if (temp != NULL)
{
temp = temp->next;
}
else
{
return NULL;
}
}

ListNode *kNode = head;
while (temp != NULL)
{
temp = temp->next;
kNode = kNode->next;
}

return kNode;
}

int main()
{
ListNode *head = new ListNode(0);
ListNode *cur = head;
for (int i = 1; i < 10; i++)
{
ListNode *newnode = new ListNode(i);
cur->next = newnode;
cur = newnode;
}

cout<<findKthToTail(head, 1)->val<<endl;
dropList(head);
}

6、反转单链表;

见博文递归与非递归反转链表

7、通过两个栈实现一个队列;

#include <iostream>
#include <stack>
using namespace std;

template<class T>
class CQueue
{
public:
void push(T& t);
void pop();
T& top();
int size();
private:
stack<T> s1;
stack<T> s2;
void gather(stack<T>& s1, stack<T>& s2);
};

template<class T> void CQueue<T>::push(T& t)
{
gather(s1, s2);
s1.push(t);
}

template<class T> void CQueue<T>::pop()
{
gather(s2, s1);
s2.pop();
}

template<class T> T& CQueue<T>::top()
{
gather(s2, s1);
T& t = s2.top();

return t;
}

template<class T> int CQueue<T>::size()
{
return s1.size() + s2.size();
}

template<class T> void CQueue<T>::gather(stack<T>& s1, stack<T>& s2)
{
while(!s2.empty())
{
s1.push(s2.top());
s2.pop();
}
}

int main()
{
CQueue<int> q;
for (int i = 0; i < 10; i++)
{
q.push(i);
}

while (q.size() > 0)
{
cout<<q.top()<<" ";
q.pop();
}
}

8、二分查找;

int binarySearch(int a[], int len, int val)
{
int begin = 0;
int end = len - 1;
int mid = (begin + end) / 2;
while (begin <= end)
{
if (a[mid] > val)
{
end = mid - 1;
}
else if(a[mid] < val)
{
begin = mid + 1;
}
else
{
return mid;
}
mid = (begin + end) / 2;
}

return -1;
}

int main()
{
int a[] = {1,2,3,4,5,6,7,8,9};
cout<<binarySearch(a, 9, 1);
}

9、快速排序

详细介绍见博文九大排序算法总结

void quickSort(int a[], int left, int right)
{
if(left >= right)
{
return;
}
int val = a[left];
int begin = left;
int end = right;
while (begin < end)
{
while(begin < end && a[end] >= val)
{
end--;
}
a[begin++] = a[end];

while(begin < end && a[begin] <= val)
{
begin++;
}
a[end--] = a[begin];
}
a[begin] = val;

quickSort(a, left, begin - 1);
quickSort(a, end + 1, right);
}

int main()
{
int a[] = {2,3,8,9,6,4,5,7,1,0};
int len = sizeof(a)/sizeof(a[0]);
quickSort(a, 0, len - 1);
for(int i = 0; i < len; i++)
{
cout<<a[i]<<" ";
}
}

10、获得一个int型的数中二进制中1的个数;

int find1count(int n)
{
int count = 0;
while (n)
{
n &= (n - 1);
count++;
}

return count;
}

11、输入一个数组，实现一个函数，让所有奇数都在偶数前面;

void RecordOddEven(int a[],int len)
{
int i = 0;
int j = len - 1;
while (i < j)
{
while (i < j && a[i] % 2 == 1)
{
i++;
}
while (i < j && a[j] % 2 == 0)
{
j--;
}

swap(a[i], a[j]);
i++;
j--;
}
}

12、判断一个字符串是否是另一个字符串的子串;

KMP算法，详见博文KMP算法字符串匹配。

#include <iostream>
using namespace std;

void prefixFun(char *pattern, int *prefun)
{
int len = 0;
while ('\0' != pattern[len])
{
len++;
}

int count = 0;
prefun[1] = 0;
for (int i = 2; i <= len; i++)
{
while (count > 0 && pattern[count] != pattern[i - 1])
{
count = prefun[count];
}

if (pattern[count] == pattern[i - 1])
{
count++;
}
prefun[i] = count;
}
}

void KMPstrMatching(char *target, char *pattern)
{
int tarLen = 0;
while ('\0' != target[tarLen])
{
tarLen++;
}

int patLen = 0;
while ('\0' != pattern[patLen])
{
patLen++;
}

int *prefun = new int[patLen + 1];
prefixFun(pattern, prefun);
int count = 0;
for (int i = 0; i < tarLen; i++)
{
while (count > 0 && pattern[count] != target[i])
{
count = prefun[count];
}

if (pattern[count] == target[i])
{
count++;
}

if (count == patLen)
{
cout<<"Patterns matching at index: "<<i - patLen + 1<<endl;
count = prefun[count];
}
}

delete [] prefun;
}

int main()
{
KMPstrMatching("abcdabababcdabcdabcabcd", "abcd");
}

13、把一个int型数组中的数字拼成一个串，这个串代表的数字最小;

先将数字转换成字符串存在数组中，在通过qsort排序，在排序用到的比较函数中，将要比较的两个字符串进行组合，如要比较的两个字符串分别是A,B，那么组合成，A+B，和B+A，在比较A+B和B+A，返回strcmp(A+B, B+A)，经过qsort这么一排序,数组就变成从小到大的顺序了，组成的数自然是最小的。

//把一个int型数组中的数字拼成一个串，是这个串代表的数组最小
#define MaxLen 10
int Compare(const void* str1,const void* str2)
{
char cmp1[MaxLen*2+1];
char cmp2[MaxLen*2+1];
strcpy(cmp1,*(char**)str1);
strcat(cmp1,*(char**)str2);

strcpy(cmp2,*(char**)str2);
strcat(cmp2,*(char**)str1);
return strcmp(cmp1,cmp2);
}
void GetLinkMin(int a[],int len)
{
char** str=(char**)new int[len];
for (int i=0;i<len;i++)
{
str[i]=new char[MaxLen+1];
sprintf(str[i],"%d",a[i]);
}

qsort(str,len,sizeof(char*),Compare);
for (int i=0;i<len;i++)
{
cout<<str[i]<<" ";
delete[] str[i] ;
}
delete[] str;
}
void GetLinkMinTest()
{
int arr[]={123,132,213,231,321,312};
GetLinkMin(arr,sizeof(arr)/sizeof(int));
}

14、输入一颗二叉树，输出它的镜像（每个节点的左右子节点交换位置）;

递归实现，只要某个节点的两个子节点都不为空，就左右交换，让左子树交换，让右子树交换。

struct NodeT
{
int value;
NodeT* left;
NodeT* right;
NodeT(int value_=0,NodeT* left_=NULL,NodeT* right_=NULL):value(value_),left(left_),right(right_){}
};

//输入一颗二叉树，输出它的镜像（每个节点的左右子节点交换位置）
void TreeClass(NodeT* root)
{
if( root==NULL || (root->left==NULL && root->right==NULL) )
return;
NodeT* tmpNode=root->left;
root->left=root->right;
root->right=tmpNode;
TreeClass(root->left);
TreeClass(root->right);
}

void PrintTree(NodeT* root)
{
if(root)
{
cout<<root->value<<" ";
PrintTree(root->left);
PrintTree(root->right);
}
}

void TreeClassTest()
{
NodeT* root=new NodeT(8);
NodeT* n1=new NodeT(6);
NodeT* n2=new NodeT(10);
NodeT* n3=new NodeT(5);
NodeT* n4=new NodeT(7);
NodeT* n5=new NodeT(9);
NodeT* n6=new NodeT(11);
root->left=n1;
root->right=n2;
n1->left=n3;
n1->right=n4;
n2->left=n5;
n2->right=n6;
PrintTree(root);
cout<<endl;
TreeClass( root );
PrintTree(root);
cout<<endl;
}

15、输入两个链表，找到它们第一个公共节点;

如果两个链表有公共的节点，那么第一个公共的节点及往后的节点都是公共的。从后往前数N个节点（N=短链表的长度节点个数）,长链表先往前走K个节点（K=长链表的节点个数-N），这时两个链表都距离末尾N个节点，现在可以一一比较了，最多比较N次，如果有两个节点相同就是第一个公共节点，否则就没有公共节点。

另一种思路是借助栈，详见博文Intersection of Two Linked List

//输入两个链表，找到它们第一个公共节点
int GetLinkLength(NodeL* head)
{
int count=0;
while (head)
{
head=head->next;
count++;
}
return count;
}

NodeL* FindFirstEqualNode(NodeL* head1,NodeL* head2)
{
if (head1==NULL || head2==NULL)
return NULL;
int len1=GetLinkLength(head1);
int len2=GetLinkLength(head2);
NodeL* longNode;
NodeL* shortNode;
int leftNodeCount;
if (len1>len2)
{
longNode=head1;
shortNode=head2;
leftNodeCount=len1-len2;
}else{
longNode=head2;
shortNode=head1;
leftNodeCount=len2-len1;
}
for (int i=0;i<leftNodeCount;i++)
{
longNode=longNode->next;
}
while (longNode && shortNode && longNode!=shortNode)
{
longNode=longNode->next;
shortNode=shortNode->next;
}
if (longNode)//如果有公共节点，必不为NULL
{
return longNode;
}
return NULL;
}

void FindFirstEqualNodeTest()
{
NodeL* head1=new NodeL(0);
NodeL* head2=new NodeL(0);
NodeL* node1=new NodeL(1);
NodeL* node2=new NodeL(2);
NodeL* node3=new NodeL(3);
NodeL* node4=new NodeL(4);
NodeL* node5=new NodeL(5);
NodeL* node6=new NodeL(6);
NodeL* node7=new NodeL(7);

head1->next=node1;
node1->next=node2;
node2->next=node3;
node3->next=node6;//两个链表相交于节点node6

head2->next=node4;
node4->next=node5;
node5->next=node6;//两个链表相交于节点node6
node6->next=node7;

NodeL* node= FindFirstEqualNode(head1,head2);
if (node)
{
cout<<node->value<<endl;
}else{
cout<<"没有共同节点"<<endl;
}
}