顺时针打印链表

void Matrix(vector
     
      
       >& num, int x1, int y1, int x2, int y2){    int n = 1;    while (x1 <= x2 && y1 <= y2)    {        int i = x1, j = y1;        for (int j = y1; j <= y2; j++)        {            num[x1][j] = n++; //right        }        if (x1 < x2)        {            for (int i = x1 + 1; i <= x2; i++)            {                num[i][y2] = n++; //down            }        }        if (x1 < x2 && y1 < y2)        {            for (int j = y2 - 1; j >= y1; j--)            {                num[x2][j] = n++; //left            }        }        if (x2 - x1 > 1 && y1 < y2)        {            for (int i = x2 - 1; i > x1; i--)            {                num[i][y1] = n++; //up            }        }        x1++, y1++, x2--, y2--;    }}
      
     

包含min函数的栈

template
     
       class StackWithMin{public:    void push(T t)    {        m_data.push(t);        if (m_min.empty() || t < m_min.top())        {            m_min.push(t);        }        else        {            m_min.push(m_min.top());        }    }    void pop()    {        assert(m_data.size() > 0 && m_min.size() > 0);        m_data.pop();        m_min.pop();    }    T top()    {        assert(m_data.size() > 0 && m_min.size() > 0);        return m_data.top();    }    T min()    {        assert(m_data.size() > 0 && m_min.size() > 0);        return m_min.top();    }    bool empty()    {        return m_data.empty();    }private:    stack
      
        m_data;    stack
       
         m_min;};
       
      
     

栈的压入弹出序列

bool isPopOrder(int num1[], int num2[], int len){    int *p = num1;    int *q = num2;    if (p != NULL && q != NULL && len > 0)    {        stack
     
       s;        while (q - num2 < len)        {            while (s.empty() || s.top() != *q)            {                if (p - num1 == len)                {                    break;                }                s.push(*p++);            }            if (s.top() != *q)            {                break;            }            s.pop();            q++;        }        if (s.empty() && q - num2 == len)        {            return true;        }    }    return false;}
     

二叉搜索树的后序遍历序列

bool verifyBST(int num[], int len){    if (num == NULL || len <= 0)    {        return false;    }    int i = 0;    while (i < len - 1)    {        if (num[i] < num[len - 1])        {            i++;        }        else        {            break;        }    }    int j = i;    while (j < len - 1)    {        if (num[j] > num[len - 1])        {            j++;        }        else        {            return false;        }    }    bool left = true;    if (i > 0)    {        left = verifyBST(num, i);    }    bool right = true;    if (i < len - 1)    {        right = verifyBST(num + i, len - i - 1);    }    return left && right;}

二叉树中和为某一值的路径

void findPath(TreeNode *root, int expectedSum, vector
     
      & path, int currentSum){    if (root == NULL)    {        return;    }    path.push_back(root->val);    currentSum += root->val;    bool isLeaf = root->left == NULL && root->right == NULL;    if (expectedSum = currentSum && isLeaf)    {        for (auto it = path.begin(); it != path.end(); it++)        {            cout<<*it<<" ";        }        cout<
      
       left, expectedSum, path, currentSum);    findPath(root->right, expectedSum, path, currentSum);    currentSum -= root->val;    path.pop_back();}void findPath(TreeNode *root, int expectedSum){    if (root == NULL)    {        return;    }    vector
       
         path;    int currentSum = 0;    findPath(root, expectedSum, path, currentSum);}
       
      
     

复杂链表的复制

ComplexListNode* copyList(ComplexListNode *head){    if (head == NULL)    {        return NULL;    }    //copy list    ComplexListNode *p = head;    while (p != NULL)    {        ComplexListNode *p2 = new ComplexListNode(p->val);        p2->next = p->next;        p->next = p2;        p = p2->next;    }    //deal with the sibling    p = head;    while (p != NULL)    {        ComplexListNode *p2 = p->next;        if (p->sib != NULL)        {            p2->sib = p->sib->next;        }        p = p2->next;    }    //separate the list    ComplexListNode *newhead = head->next;    ComplexListNode *pre1 = head;    ComplexListNode *pre2 = newhead;    p = newhead->next;    while (p != NULL)    {        pre1->next = p;        pre1 = p;        pre2->next = p->next;        pre2 = p->next;        p = p->next->next;    }    pre1->next = NULL;    return newhead;}

二义搜索树与双向链表

void convertHelp(TreeNode *root, TreeNode *&last){    if (root == NULL)    {        return;    }    TreeNode *cur = root;    if (cur->left != NULL)    {        convertHelp(root->left, last);    }    cur->left = last;    if (last != NULL)    {        last->right = cur;    }    last = cur;    if (cur->right != NULL)    {        convertHelp(cur->right, last);    }}TreeNode* convert(TreeNode *root){    TreeNode *last = NULL;    convertHelp(root, last);    TreeNode *p = last;    while (p != NULL &&p->left != NULL)    {        p = p->left;    }    return p;}

字符串的排列

void perHelp(char *str1, char *str2) {    if (*str2 == '\0')    {        cout<
     
      <
     

全排列

int n[10] = {0,1,2,3,4,5,6,7,8,9};void permutation(int num[], bool flag[], int len, int pos){    for (int i = 0; i < len; i++)    {        if (!flag[i])        {            num[pos] = n[i];            flag[i] = true;            if (pos == len - 1)            {                print(num, len);            }            else            {                permutation(num, flag, len, pos + 1);            }            flag[i] = false;        }    }}

组合

/* 算法说明：从n个数中选m个数，可以分解为以下两步: * （1）首先从n个数中选取编号最大的数，然后在剩下的n-1个数中选取m-1个数，直到从n-(m-1)个数中选取1个数为止。 * （2）从n个数中选取编号次小的一个数，继续执行第（1）步，直到当前可选编号最大的数为m。 *//* 求从a[n]中选取m个数的可能组合。数组a[n]表示候选集；b[m]用来存储当前组合中的某个元素， * 这里存储的是这个元素在a[]中的下标；常量M表示满足条件的一个组合中元素的个数，M=m，M只用来输出结果 */void combine(int a[], int b[], int n, int m, int M){    for (int i = n; i >= m; i--)    {        b[m - 1] = a[i - 1];        if (m > 1)        {            combine(a, b, i - 1, m - 1, M);        }        else        {            for (int j = M - 1; j >= 0; j--)            {                cout<
     
      <<" ";            }            cout<