leetcode递归算法题总结

递归本质是找重复的子问题

本章目录

1.汉诺塔
2.合并两个有序链表
3.反转链表
4.两两交换链表中的节点
5.Pow(x,n)

1.汉诺塔

汉诺塔
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//面试写法
class Solution {
public:void hanota(vector<int>& a, vector<int>& b, vector<int>& c) {dfs(a,b,c,a.size());}void dfs(vector<int>& a, vector<int>& b, vector<int>& c,int n){if(n==0) return;if(n==1){c.push_back(a.back());a.pop_back();return;}dfs(a,c,b,n-1);c.push_back(a.back());a.pop_back();dfs(b,a,c,n-1);}
};

//笔试写法
class Solution {
public:void hanota(vector<int>& A, vector<int>& B, vector<int>& C) {C =A;}
};

2.合并两个有序链表

合并两个有序链表
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/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {if(list1 == nullptr) return list2;if(list2 == nullptr) return list1;if(list1->val<=list2->val) {list1->next = mergeTwoLists(list1->next,list2);return list1;}else {list2->next = mergeTwoLists(list1,list2->next);return list2;}}
};

3.反转链表

反转链表
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/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* reverseList(ListNode* head) {if(head == nullptr || head->next == nullptr) return head;ListNode* newHead = reverseList(head->next);head->next->next = head;head->next = nullptr;return newHead;}
};

4.两两交换链表中的节点

两两交换链表中的节点
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/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* swapPairs(ListNode* head) {//递归if(head == nullptr || head->next == nullptr) return head;auto tmp = swapPairs(head->next->next);auto ret = head->next;head->next->next = head;head->next = tmp;return ret;}
};

class Solution {
public://模拟 循环 迭代ListNode* swapPairs(ListNode* head) {if(head == nullptr || head->next == nullptr) return head;ListNode* ret = new ListNode(0);ListNode* prev = ret;prev->next = head;ListNode* cur = prev->next,*next = cur->next,*nnext = next->next;while(cur && next){//交换节点prev->next = next;next->next = cur;cur->next = nnext;//更新节点prev = cur;cur = nnext;if(cur) next = cur->next;if(next) nnext = next->next;}cur = ret->next;delete ret;return cur;}
};

5.Pow(x,n)

Pow(x,n)
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class Solution {
public:double myPow(double x, int n) {//递归+快速幂return n<0? 1.0/pow(x,-(long long)n):pow(x,n);}double pow(double x,long long n){if(n==0) return 1;double tmp = pow(x,n/2);return n%2==0? tmp*tmp:tmp*tmp*x;}
};