目录

109:旋转字符串

110:合并k个已排序的链表

111:滑雪

---

109:旋转字符串

题目链接:旋转字符串_牛客题霸_牛客网 (nowcoder.com)

题目:

题解:

class Solution {
public:bool solve(string A, string B) {int n=A.size();if(n!=B.size()) return false;for(int i=0;i<n;i++){int j=0;if(A[i]==B[j]){int left=i,right=j;while(A[left]==B[right]){left++;left%=n;right++;if(right==n-1) return true;}}}return false;  }
};

110:合并k个已排序的链表

题目链接:合并k个已排序的链表_牛客题霸_牛客网 (nowcoder.com)

题目:

题解:

解法1：建小堆，将所有节点放进堆中，再从堆顶逐个取出组成链表。

/*** struct ListNode {*	int val;*	struct ListNode *next;*	ListNode(int x) : val(x), next(nullptr) {}* };*/
#include <queue>
class Solution {
public://template<class T>class compare{public:bool operator()(const ListNode* x, const ListNode* y){return x->val > y->val;}};ListNode* mergeKLists(vector<ListNode*>& lists) {ListNode* newhead=new ListNode(0);priority_queue<ListNode*, vector<ListNode*>, compare> heap;for(int i=0;i<lists.size(); i++){ListNode* cur=lists[i];while(cur){heap.push(cur);cur=cur->next;}}ListNode* cur=newhead;while(!heap.empty()){cur->next=heap.top();heap.pop();cur=cur->next;}cur->next=nullptr;return newhead->next;}
};

解法2：使用 std::sort 对所有链表节点指针调用compare()进行排序。

/*** struct ListNode {*	int val;*	struct ListNode *next;*	ListNode(int x) : val(x), next(nullptr) {}* };*/
#include <queue>
class Solution {
public://template<class T>class compare{public:bool operator()(const ListNode* x, const ListNode* y){return x->val > y->val;}};ListNode* mergeKLists(vector<ListNode*>& lists) {ListNode* newhead=new ListNode(0);vector<ListNode*> list;for(int i=0;i<lists.size(); i++){ListNode* cur=lists[i];while(cur){list.push_back(cur);cur=cur->next;}}ListNode* cur=newhead;sort(list.begin(), list.end(), compare());for(int i=list.size()-1;i>=0;i--){cur->next=list[i];cur=cur->next;}cur->next=nullptr;return newhead->next;}
};

 解法3：因为各个小链表是排好序的，可以先将各个链表头节点加入堆中，从堆顶中找小的插入新链表中再pop()后，将堆顶节点的下一节点重新插入堆。

/*** struct ListNode {*	int val;*	struct ListNode *next;*	ListNode(int x) : val(x), next(nullptr) {}* };*/
class Solution {
public://template<class T>class compare{public:bool operator()(const ListNode* x, const ListNode* y){return x->val > y->val;}};ListNode* mergeKLists(vector<ListNode*>& lists) {ListNode* newhead=new ListNode(0);priority_queue<ListNode*, vector<ListNode*>, compare> heap;for(auto head:lists){if(head!=nullptr){heap.push(head);}}ListNode* cur=newhead;while(heap.size()){ListNode* t=heap.top();heap.pop();cur->next=t;cur=cur->next;if(t->next!=nullptr){heap.push(t->next);}}cur->next=nullptr;return newhead->next;}
};

解法4：利用归并排序思想，将链表一层一层向下划分直到为单个链表，再向上合并同时排序链表节点。

class Solution {public://两个有序链表合并函数ListNode* Merge2(ListNode* pHead1, ListNode* pHead2) {//一个已经为空了，直接返回另一个if (pHead1 == NULL)return pHead2;if (pHead2 == NULL)return pHead1;//加一个表头ListNode* head = new ListNode(0);ListNode* cur = head;//两个链表都要不为空while (pHead1 && pHead2) {//取较小值的节点if (pHead1->val <= pHead2->val) {cur->next = pHead1;//只移动取值的指针pHead1 = pHead1->next;} else {cur->next = pHead2;//只移动取值的指针pHead2 = pHead2->next;}//指针后移cur = cur->next;}//哪个链表还有剩，直接连在后面if (pHead1)cur->next = pHead1;elsecur->next = pHead2;//返回值去掉表头return head->next;}//划分合并区间函数ListNode* divideMerge(vector<ListNode*>& lists, int left, int right) {if (left > right)return NULL;//中间一个的情况else if (left == right)return lists[left];//从中间分成两段，再将合并好的两段合并int mid = (left + right) / 2;return Merge2(divideMerge(lists, left, mid), divideMerge(lists, mid + 1,right));}ListNode* mergeKLists(vector<ListNode*>& lists) {//k个链表归并排序return divideMerge(lists, 0, lists.size() - 1);}
};

111:滑雪

题目链接:滑雪_牛客题霸_牛客网 (nowcoder.com)

题目:

题解:

dfs+记忆化搜索

#include <iostream>
using namespace std;const int N=110;
int m=0,n=0;
int board[N][N]={0};
int vis[N][N]={0};int dx[4]={0,0,-1,1};
int dy[4]={-1,1,0,0};int dfs(int i, int j)
{if(vis[i][j]) return vis[i][j];int len=1;for(int k=0;k<4;k++){int x=i+dx[k], y=j+dy[k];if(x>=0 && y>=0 && x<n && y<m && board[x][y] > board[i][j]){len=max(len, 1+dfs(x, y));}}vis[i][j]=len;return len;
}int main() 
{cin>>n>>m;for(int i=0;i<n;i++){for(int j=0;j<m;j++){cin>>board[i][j];}}int ret=0;for(int i=0;i<n;i++){for(int j=0;j<m;j++){ret=max(ret, dfs(i,j));}}cout<<ret<<endl;return 0;
}