1. 链表

链表逆转

 class Solution {
public:
ListNode* reverseList(ListNode* head) {
ListNode* prev = NULL;
while (head != NULL) {
ListNode* next = head->next;
head->next = prev;
prev = head;
head = next;
}
return prev;
}
};

findMiddle

树的{前、中、后、层}序遍历

堆排序

快排

字符串的反转

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