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每日一题 链表定义 cpp struct ListNode{ int val; ListNode *next; ListNode(int x) : val(x),next(NULL) {} } python class ListNode: def __init__(self, val, next = None): self.val = val self.next = next 移除链表元素 https://leetcode.cn/problems/remove-linked-list-elements/ cpp /** * 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* removeElements(ListNode* head, int val) { while(head != NULL && head->val == val){ ListNode* temp = head; head = head-> next; delete temp; } ListNode* cur = head; while(cur != NULL && cur->next != NULL){ if(cur->next->val == val){ ListNode* temp = cur->next; cur->next = cur->next->next; delete temp; }else{ cur = cur->next; } } return head; } }; python ...

每日一题 滑动窗口 leetcode 209 长度最小的子数组 https://leetcode.cn/problems/minimum-size-subarray-sum/ 代码： cpp class Solution { public: int minSubArrayLen(int target, vector<int>& nums) { int sum = 0; int j = 0; int res = nums.size() + 10; for(int i = 0; i < nums.size(); i++ ){ sum += nums[i]; while(sum >= target ){ res = min(res, i - j + 1); sum -= nums[j++]; } } res = (res == nums.size() + 10 ? 0 : res); return res; } }; python: class Solution: def minSubArrayLen(self, target: int, nums: List[int]) -> int: res = len(nums) + 10 sum = 0 j = 0 for i in range(len(nums)): sum += nums[i] while sum >= target: res = min(res, i - j + 1) sum -= nums[j] j += 1 if res == len(nums) + 10: res = 0 return res leetcode 904 水果成篮 https://leetcode.cn/problems/fruit-into-baskets/ 代码： cpp: class Solution { public: int totalFruit(vector<int>& fruits) { int res = 0; unordered_map<int, int> cnt; int j = 0; for(int i = 0; i < fruits.size(); i++){ cnt[fruits[i]]++; while(cnt.size() > 2){ auto it = cnt.find(fruits[j]); it->second--; if(it->second == 0) cnt.erase(fruits[j]); j++; } res = max(res, i - j + 1); } return res; } }; python ...

每日一题 双指针算法 leetcode 27 移出元素 https://leetcode.cn/problems/remove-element/ 代码: cpp: class Solution { public: int removeElement(vector<int>& nums, int val) { int fast = 0; int slow = 0; for (fast = 0; fast < nums.size(); fast++){ if(nums[fast] != val){ nums[slow++] = nums[fast]; } } return slow; } }; python: class Solution: def removeElement(self, nums: List[int], val: int) -> int: slow, fast = 0, 0 for fast in range(len(nums)): if nums[fast] != val: nums[slow] = nums[fast] slow += 1 return slow leetcode 26.删除有序数组中的重复项 https://leetcode.cn/problems/remove-duplicates-from-sorted-array/ 代码 cpp: class Solution { public: int removeDuplicates(vector<int>& nums) { int fast; int slow = 0; for(fast = 1; fast < nums.size(); fast++){ if(nums[fast] != nums[slow]){ slow++; nums[slow] = nums[fast]; } } return slow + 1; } }; python: ...

每日一题 二分查找 leetcode 704 二分查找 https://leetcode.cn/problems/binary-search/ 题解： CPP: class Solution { public: int search(vector<int>& nums, int target) { int left = 0; int right = nums.size() - 1; while(left < right){ int middle = left + right >> 1; if(nums[middle] >= target){ right = middle; }else{ left = middle + 1; } } if(nums[left] == target) return left; return -1; } }; class Solution { public: int search(vector<int>& nums, int target) { int left = 0; int right = nums.size() - 1; while(left < right){ int middle = left + right + 1 >> 1; if(nums[middle] > target){ right = middle - 1; }else{ left = middle; } } if(nums[left] == target) return left; return -1; } }; python: class Solution: def search(self, nums: List[int], target: int) -> int: left, right = 0,len(nums) - 1 while left < right: middle = left + right + 1 >> 1 if nums[middle] > target: right = middle - 1 elif nums[middle] <= target: left = middle if nums[left] == target: return left return -1 class Solution: def search(self, nums: List[int], target: int) -> int: left, right = 0,len(nums) - 1 while left < right: middle = left + right >> 1 if nums[middle] >= target: right = middle elif nums[middle] < target: left = middle + 1 if nums[left] == target: return left return -1 leetcode 35.搜索插入位置 https://leetcode.cn/problems/search-insert-position/ ...

十三、STL函数 beg为序列的初始地址 end为序列的尾地址 accumulate accumulate(beg, end, init) 复杂度: O(N) 作用：对一个序列的元素求和 init为对序列元素求和的初始值 返回值类型：与init 相同 基础累加求和： ...

十一、array数组 11.1 介绍 头文件： #include<array> array是C++11新增的容器，效率与普通数据相差无几，比vector效率要高，自身添加了一些成员函数。 和其它容器不同，array 容器的大小是固定的，无法动态的扩展或收缩，只允许访问或者替换存储的元素。 ...

六、map 6.1 介绍 映射类似于函数的对应关系，每一个x对应一个y,而map是每个键对应一个值。和python的字典非常相似。 //头文件 #include<map> //初始化定义 map<string,string> mp; map<string,int> mp; map<int,node> mp;//node是结构体类型 map特性：map会按照键的顺序从小到大自动排序，键的类型必须可以比较大小 ...