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 list
void 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;
}
};