LeetCode LRU Cache & Insertion Sort List

LRU Cache

 Total Accepted: 16727 Total Submissions: 120052My Submissions

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

这一题需要O(log n)的复杂度才能过，于是可以借助map的红黑树。

trick是map里存的是list的iterator.

用list记录顺序。list大概是类似双向队列的东西。然后begin() end() front() back()竟然搞混了。

class LRUCache{    struct CacheNode {int key;int val;CacheNode(int a, int b) : key(a), val(b) {}};public:LRUCache(int capacity) {cap = capacity;}int get(int key) {if (cacheMap.find(key) != cacheMap.end()) {    touch(key);return cacheList.begin()->val;}else return -1;}int set(int key, int val) {if (cacheMap.find(key) != cacheMap.end()) {    touch(key);cacheMap[key]->val = val;}else {if (cacheList.size() >= cap) {cacheMap.erase(cacheList.back().key);cacheList.pop_back();}cacheList.push_front(CacheNode(key, val));cacheMap[key] = cacheList.begin();}}// CacheNode key already exists, so transfer it to begin of the listvoid touch(int key) {    // solution 1  //  cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);    // solution 2    cacheList.push_front(*cacheMap[key]);    cacheList.erase(cacheMap[key]);        cacheMap[key] = cacheList.begin();}int cap;map<int, list<CacheNode>::iterator> cacheMap;list<CacheNode> cacheList;};

Insertion Sort List

 Total Accepted: 20174 Total Submissions: 79737My Submissions

Sort a linked list using insertion sort.

指针类型的插入排序。

所以要用指针记录已经排好的头和尾。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode *insertionSortList(ListNode *head) {        if (head == NULL) return head;ListNode *ans = head;ListNode *tail = head;ListNode *p, *tmp;while (tail->next) {int val = tail->next->val;if (val < ans->val) {//insert to the front tmp = tail->next;tail->next = tmp->next;tmp->next = ans;ans = tmp;}else {for (p = ans; p->next->val <= val && p != tail; p = p->next);if (p != tail) {tmp = tail->next;tail->next = tmp->next;tmp->next = p->next;p->next = tmp;}else tail = tail->next;}return ans;       }};