剑指offer:字符串的排列class Solution {public:    vector<string> Permutation(string str) {        if(str.empty())            return {};        sort(str.begin(),str.end());        vector<string> answer;        answer.push_back(str);        while(next_permutation(str.begin(),str.end()))        {            answer.push_back(str);        }        return answer;     }};

剑指offer08: 跳台阶class Solution {public:    int jumpFloor(int number) {        int first=1,second=1,flag=0;        if(number==1)            return 1;        else if(number==2)            return 2;        else        {            for(int i=1;i<number;i++)            {                flag=first+second;                first = second;                second = flag;            }        }        return flag;    }};

剑指offer:变态跳台阶class Solution {public:    int jumpFloorII(int number) {        //通过移位计算2的次方        return 1<<(number-1);      }};

剑指offer: 链表中环的入口节点class Solution {public:    ListNode* EntryNodeOfLoop(ListNode* pHead)    {        ListNode* fast=pHead,*low=pHead;        while(fast&&fast->next)        {            fast=fast->next->next;            low=low->next;            if(fast==low)                break;        }        if(!fast||!fast->next)            return NULL;        low=pHead; //low从链表的头出发        while(fast!=low)        {            fast=fast->next;            low=low->next;        }        return low;    }};

剑指offer: 裴波那契数列class Solution {public:    int Fibonacci(int n) {        if(n==0)            return 0;        else if(n==1)            return 1;        int answer=0;        int first=0,second=1;        for(int i =2;i<=n;i++)        {            answer = first+second;            first = second;            second = answer;        }        return answer;    }};

剑指offer ：删除链表中重复的结点class Solution {public:    ListNode* deleteDuplication(ListNode* pHead)    {        if (pHead==NULL)            return NULL;        if (pHead!=NULL && pHead->next==NULL)            return pHead;                         ListNode* current;                 if ( pHead->next->val==pHead->val){            current=pHead->next->next;            while (current != NULL && current->val==pHead->val)                current=current->next;            return deleteDuplication(current);                             }                 else {            current=pHead->next;            pHead->next=deleteDuplication(current);            return pHead;        }        }};

剑指offer 表示数值的字符串class Solution {public:    bool isNumeric(char* str) {        // 标记符号、小数点、e是否出现过        bool sign = false, decimal = false, hasE = false;        for (int i = 0; i < strlen(str); i++) {            if (str[i] == e || str[i] == E) {                if (i == strlen(str)-1) return false; // e后面一定要接数字                if (hasE) return false;  // 不能同时存在两个e                hasE = true;            } else if (str[i] == + || str[i] == -) {                // 第二次出现+-符号，则必须紧接在e之后                if (sign && str[i-1] != e && str[i-1] != E) return false;                if (!sign && i > 0 && str[i-1] != e && str[i-1] != E) return false;                sign = true;            } else if (str[i] == .) {                if (hasE || decimal) return false;                decimal = true;            } else if (str[i] < 0 || str[i] > 9) // 不合法字符                return false;        }        return true;    }};

剑指offer: 正则表达式匹配class Solution {public:    bool match(char* str, char* pattern)    {        if (*str ==  && *pattern == )            return true;        if (*str !=  && *pattern == )            return false;        //if the next character in pattern is not *        if (*(pattern+1) != *)        {            if (*str == *pattern || (*str !=  && *pattern == .))                return match(str+1, pattern+1);            else                return false;        }        //if the next character is *        else        {            if (*str == *pattern || (*str !=  && *pattern == .))                return match(str, pattern+2) || match(str+1, pattern);            else                return match(str, pattern+2);        }    }};

剑指offer: 字符流中第一个不重复的字符class Solution{public:  //Insert one char from stringstream    void Insert(char ch)    {          vec.push_back(ch);          mapdata[ch]++;    }  //return the first appearence once char in current stringstream    char FirstAppearingOnce()    {          vector<int>::iterator it;          for(it=vec.begin();it!=vec.end();it++)          {               if(mapdata[*it]==1)                    return *it;          }      return #;    }         map<char,int> mapdata;         vector<int> vec;};

剑指offer 替换空格class Solution {public:    void replaceSpace(char *str,int length) {        for(int i=0;i<length;i++)        {            if(str[i]== ) //如果存在空格            {                for(int j=length+1;j>=i+3;j--)  //后面的元素向后移动两位                    str[j]=str[j-2];                str[i] = %;                str[i+1] = 2;                str[i+2] = 0;                length+=2;            }        }    }};

剑指offer: 二维数组查找class Solution {public:    bool Find(int target, vector<vector<int> > array) {        int lie = array[0].size();        int hang = array.size();        int i,j;        for(i = 0;i<hang;i++)        {            for(j=0;j<lie;j++)            {                if(array[i][j]==target)                    return true;            }            j=0;        }        return false;            }};

剑指offer 3 : 从尾到头打印链表class Solution {public:    vector<int> printListFromTailToHead(ListNode* head) {        vector<int> nums;        while(head!=NULL)        {            nums.push_back(head->val);            head=head->next;        }        reverse(nums.begin(),nums.end());        return nums;            }};

剑指offer14: 调整数组顺序使奇数位于偶数前面class Solution {public:    void reOrderArray(vector<int> &array) {        vector<int> jishu;        for(int i=0;i<array.size();i++)        {            if(array[i]%2!=0)                jishu.push_back(array[i]);        }        for(int i=0;i<array.size();i++)        {            if(array[i]%2==0)                jishu.push_back(array[i]);        }        array=jishu;    }};

剑指offer42: 和为S的两个数字class Solution {public:    vector<int> FindNumbersWithSum(vector<int> array,int sum) {        int flag=INT_MAX;        vector<int> b(2,-1);        for(int i=0;i<array.size();i++)            for(int j=i+1;j<array.size();j++)            {                if(array[i]+array[j]==sum)                {                    if(array[i]*array[j]<flag)                    {                        b[0]=array[i];                        b[1]=array[j];                        flag=b[0]*b[1];                    }                                    }            }        if(b[0]+b[1]!=-2)            return b;        else            return {};            }};