退役老人现在连leetcode都不会做了 = =
今天早上做了leetcode第三题题目看错了，加上比赛中间还在调投稿的实验，一心二用直接gg
总结下教训就是 本渣现在做题连题目都看不清就开始做。开始写题之前应当把样例过一遍，然后自己再造1-2个例子，然后再开始做
A题：统计素数的个数(素数筛或者sqrt(n)判断都可以)，然后分别计算count!
class Solution {
public:
int numPrimeArrangements(int n) {
vector<int> has(n + 5, 0);
int cntPrime = 0;
const int MOD = 1e9 + 7;
for(int i = 2; i <= n; ++i) {
if(has[i] == 0) cntPrime ++;
for(int j = i * 2; j <= n; j += i) {
has[j] = 1;
}
}
int result = 1;
for(int i = 1; i <= n - cntPrime; ++i) {
result = 1ll * i * result % MOD;
}
for(int i = 1; i <= cntPrime; ++i) {
result = 1ll * i * result % MOD;
}
return result;
}
};B题：滚动求和
class Solution {
public:
int dietPlanPerformance(vector<int>& calories, int k, int lower, int upper) {
long long sum = 0;
for(int i = 0; i < k; ++i) {
sum += calories[i];
}
int result = 0;
for(int i = k - 1, len = calories.size(); i < len; ++i) {
if(sum < lower) result --;
else if(sum > upper) result ++;
if(i != len - 1) {
sum -= calories[i - k + 1];
sum += calories[i + 1];
}
}
return result;
}
};C题：我之前把题意看成整个字符串满足是回文，这个复杂很多。。。。
正确解法是判断区间的每种字母个数，我们知道偶数是可以直接配对的(放对称位置就好了)，统计字母个数为奇数的，奇数的需要改一半就好了
class Solution {
private:
vector<int> has[26];
int get(int id, int fr, int to) {
if(fr > to) return 0;
if(fr == 0) return has[id][to];
else return has[id][to] - has[id][fr - 1];
}
public:
vector<bool> canMakePaliQueries(string s, vector<vector<int>>& queries) {
int slen = s.length();
for(int i = 0; i < 26; ++i) has[i].clear();
/***statistic character count*****/
for(int i = 0; i < 26; ++i) {
for(int j = 0; j < slen; ++j) {
has[i].push_back(0);
}
}
for(int i = 0; i < slen; ++i) {
has[s[i] - 'a'][i] ++;
}
for(int i = 0; i < 26; ++i) {
for(int j = 1; j < slen; ++j) {
has[i][j] += has[i][j-1];
}
}
vector<bool> result;
for(int i = 0, len = queries.size(); i < len; ++i) {
int left = queries[i][0]; int right = queries[i][1]; int k = queries[i][2];
int cnt = 0;
for(int j = 0; j < 26; ++j) {
int t1 = get(j, left, right);
if(t1 % 2 == 1) {
cnt ++;
}
}
cnt /= 2;
if(cnt <= k) result.push_back(true);
else result.push_back(false);
}
return result;
}
};D题：基本思想就是暴力查询，首先我们知道，字符串中的字母排列前后和出现次数都是无所谓的(除了puzzle的第一个字母的问题)，这个是可以进行处理的
除此之外有两种加速的方法
方法1: 位压缩为26位，直接通过数位判断word是否是满足puzzle
方法2: 对word建立字典树，让puzzle在字典树中进行dfs，这里每个puzzle大约会有7!的查询复杂度
下面两个代码分别对应这两种方案，我是直接discuss的大佬里面爬下来了
class Solution {
vector<int> base;
void born(vector<string>& words){
for(auto& s: words){
set<char> tmp;
int bit = 0;
for(auto c:s){
tmp.insert(c);
bit = bit | (1<<(c-'a'));
}
if(tmp.size() >7)continue;
base.push_back(bit);
}
}
public:
vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {
vector<int> ans;
born(words);
for(auto& s: puzzles){
int num = 0;
int bit = 0;
for(auto c: s){
bit = bit | (1<<(c-'a'));
}
int firstBit = 1 << (s[0] - 'a');
for(auto v: base){
if((v & bit) == v && ((firstBit & v) == firstBit)){
num++;
}
}
ans.push_back(num);
}
return ans;
}
};const int ALPHABET_SIZE = 26;
/* The structure of a trie node */
struct TrieNode
{
struct TrieNode* children[ALPHABET_SIZE];
int count = 0;
};
/* Creates a new trie node and returns the pointer */
struct TrieNode* getNode()
{
struct TrieNode* newNode = new TrieNode;
for(int i=0; i<ALPHABET_SIZE; i++)
newNode->children[i] = nullptr;
newNode->count = 0;
return newNode;
}
/* Inserts the given string to the collection */
void insert(struct TrieNode* root, string str)
{
struct TrieNode* pCrawl = root;
for(int i=0; i<str.length(); i++)
{
int index = str[i]-'a';
if(!pCrawl->children[index])
pCrawl->children[index] = getNode();
pCrawl = pCrawl->children[index];
}
pCrawl->count = (pCrawl->count + 1);
}
/* Returns the count of strings which are valid */
int search(struct TrieNode* root, string str, bool firstSeen, char firstLetter)
{
if(!root)
return 0;
int count = 0;
if(firstSeen)
count += root->count;
for(int i=0; i<str.length(); i++)
{
int index = str[i] - 'a';
if(str[i] == firstLetter)
count += search(root->children[index], str, true, firstLetter);
else
count += search(root->children[index], str, firstSeen, firstLetter);
}
return count;
}
class Solution
{
public:
vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles);
};
vector<int> Solution :: findNumOfValidWords(vector<string>& words, vector<string>& puzzles)
{
struct TrieNode* root = getNode();
for(auto str : words)
{
set<char> temp;
temp.insert(str.begin(), str.end());
string sorted = "";
for(auto ele : temp)
sorted += ele;
insert(root, sorted);
}
vector<int> count;
for(auto puzzle : puzzles)
{
char firstLetter = puzzle[0];
sort(puzzle.begin(), puzzle.end());
count.push_back(search(root, puzzle, false, firstLetter));
}
return count;
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