206. 反转链表

class Solution {

    public:

        ListNode* reverseList(ListNode* head) {

            ListNode* prev = nullptr;

            ListNode* current = head;

            while(current != nullptr){

                ListNode* tempNext = current->next;

                current->next = prev;

                prev = current;

                current = tempNext;

            }

            return prev;

        }

};

92. 反转链表 II

class Solution {

    public ListNode reverseBetween(ListNode head, int left, int right) {

        int count = 1;

        ListNode current = head;

        ListNode previous = null;

        ListNode tempNext = null;

        ListNode leftBreak = null;

        ListNode reverseTail = null;

        while(count <= right){

            tempNext = current.next;

            if(left == count){

                leftBreak = previous;

                reverseTail = current;

            } else if(left < count){

                current.next = previous;

            }

            previous = current;

            current = tempNext;

            count++;

        }

        reverseTail.next = current;

        if(leftBreak != null){

            leftBreak.next = previous;

         } else {

            return previous;

        }

        return head;

    }

}

二分查找边界问题

查找最接近的数

二分法避免死循环

不存在会返回比目标数字大的，因为判断当前mid位置的数字<num时，最后一次left = mid + 1，判断当前mide位置的数字>num时，right = mide，所以是大于目标数字。

最靠右的：

function binarySearch(num, nums) {
    let left = 0, right = nums.length - 1;
    while(true) {
        if (left == right) {
        return left;
        }
        let mid = left + Math.floor((right - left) / 2);
        if (nums[mid] < num) {
            left = mid + 1;
        } else if (nums[mid] == num && nums[mid + 1] == num) {
            left = mid+1;
        } else {
            right = mid;
        }
    }
}

• 1、效率；
• 2、公平

class LRUCache {
    private int capacity;
    private int n;
    private DoubleLinkedList pHead,pTail;
    private DoubleLinkedList[] hash;

    private class DoubleLinkedList{
        int key, val;
        DoubleLinkedList prev, next;

        public DoubleLinkedList(int key, int val){
            this.key = key;
            this.val = val;
            this.prev = null;
            this.next = null;
        }
    }

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.n = 0;
        hash = new DoubleLinkedList[10001];
        pHead = new DoubleLinkedList(-1,0);
        pTail = new DoubleLinkedList(-2,0);
        pHead.next = pTail;
        pTail.prev = pHead;
    }
    
    public int get(int key) {
        DoubleLinkedList node = hash[key];
        if(node==null){
            return -1;
        }
        moveFront(node);
        return node.val;
    }
    
    public void put(int key, int value) {
        DoubleLinkedList node = hash[key];
        if(node == null && n < capacity){
            node = new DoubleLinkedList(key,value);
            hash[key] = node;
            addFront(node);
            n++;
            return;
        }
        if(node ==null && n==capacity){
            node = pTail.prev;
            hash[node.key] = null;
            hash[key] = node;
        }
        node.key = key;
        node.val = value;
        moveFront(node);
    }

    private void moveFront(DoubleLinkedList node){
        node.prev.next = node.next;
        node.next.prev = node.prev;
        addFront(node);
    }

    private void addFront(DoubleLinkedList node){
        node.prev = pHead;
        node.next = pHead.next;
        pHead.next.prev = node;
        pHead.next = node;
    }
}

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache obj = new LRUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        int count = 0;
        ListNode current = head;
        ListNode previous = null;
        ListNode newCurrent  = current;
        ListNode leftBreak = null,reverseTail = head, reverseHead = null;
        while(true){
            count ++;
            if(count == k){
                reverseHead = current;
                current = reverseTail;
                previous = null;
                while (previous != reverseHead){
                    newCurrent = current.next;
                    current.next = previous;
                    previous = current;
                    current = newCurrent;
                }
                if(leftBreak == null){
                    head = reverseHead;
                }else{
                    leftBreak.next = reverseHead;
                }
                leftBreak = reverseTail;
                reverseTail.next = current;
                reverseTail = current;
                count = 0;
            }else{
                current = current.next;
            }
            if(current == null){
                break;
            }

        }
        return head;
    }
}

1. 二分

2. 2.考点：