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/*o(n)的堆化方法*/

void myjust(vector<int>& A,int i)
{
int l=i*2+1;
int r=i*2+2;
int minn=i;
if(l<A.size()&&A[l]<A[minn])
minn=l;
if(r<A.size()&&A[r]<A[minn])
minn=r;
if(minn!=i)
{
swap(A[minn],A[i]);
myjust(A,minn);
}
}

void heapify(vector<int> &A) {
int lens=A.size();
if(lens==0) return;
for(int i=lens/2;i>=0;i--)
{
myjust(A,i);
}
}

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求sqrt(double x)

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鍵樹class Trie { public:struct TrieNode {bool isEnd;map<char, TrieNode*> myMap;TrieNode(){isEnd = false;}};TrieNode* root;/** Initialize your data structure here. */Trie() {root = new TrieNode();}/** Inserts a word into the trie. */void insert(string word) {TrieNode* current = root;for (int i = 0; i < word.length(); i++){if (current->myMap.find(word[i]) != current->myMap.end()){current = current->myMap[word[i]];if (i == word.length() - 1)current->isEnd = true;}else{TrieNode* tmp = new TrieNode();tmp->isEnd = (i == word.length() - 1 ? true : false);current->myMap[word[i]] = tmp;current = tmp;}}}/** Returns if the word is in the trie. */bool search(string word) {TrieNode* current = root;for (int i = 0; i < word.length(); i++){if (current->myMap.find(word[i]) != current->myMap.end()){current = current->myMap[word[i]];}else{return false;}}return current->isEnd == true ? true : false;}/** Returns if there is any word in the trie that starts with the given prefix. */bool startsWith(string prefix) {TrieNode* current = root;for (int i = 0; i < prefix.length(); i++){if (current->myMap.find(prefix[i]) != current->myMap.end()){current = current->myMap[prefix[i]];}else{return false;}}return true;} };

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//尋找前k大小的數(shù)

int getIdx(vector<int>& nums,int left,int right) {int tmp=nums[left];while(left<right){while (left<right&&nums[right]>=tmp)right--;nums[left]=nums[right];while (left<right&&nums[left]<tmp)left++;nums[right]=nums[left];}nums[left]=tmp;return left; }vector<int> GetLeastNumbers_Solution(vector<int> input, int k) {vector<int> ans;if(input.size()<k||k<=0||input.size()==0){return ans;}int idx=-1;int left=0,right=input.size()-1;while (idx!=k-1){idx=getIdx(input,left,right);if(idx>k-1){right=idx-1;}else{left=idx+1;}}for(int i=0;i<k;i++){ans.push_back(input[i]);}return ans; }

　　

　　

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轉(zhuǎn)載于:https://www.cnblogs.com/wuxiangli/p/6089937.html

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