1. 排序数组
. - 力扣（LeetCode）
class Solution {void heapSort(vector<int>& nums){for(int i=0;i<nums.size();i++){heapInsert(nums,i);}int l=nums.size();while(l){swap(nums[0],nums[l-1]);l--;heapIfy(nums,0,l);}}void heapInsert(vector<int>& nums,int i){while(nums[i]>nums[(i-1)/2]){swap(nums[i],nums[(i-1)/2]);i=(i-1)/2;}}void heapIfy(vector<int>& nums,int i,int size){int l=2*i+1;while(l<size){int best=((l+1)<size)&&(nums[l+1]>nums[l])?l+1:l;if(nums[best]<nums[i]){best=i;}if(best==i){break;}swap(nums[i],nums[best]);i=best;l=2*i+1;}}
public:vector<int> sortArray(vector<int>& nums) {heapSort(nums);return nums;}
};2. 寻找峰值
. - 力扣（LeetCode）
class Solution {
public:int findPeakElement(vector<int>& nums) {int l=0,r=nums.size()-1;if(nums.size()==1){return 0;}if(nums[l]>nums[l+1]){return l;}if(nums[r]>nums[r-1]){return r;}int ans=-1;while(l<=r){int m=l+((r-l)>>1);if(nums[m+1]>nums[m]){l=m+1;}else if(nums[m-1]>nums[m]){r=m-1;}else{ans=m;break;}}return ans;}
};3. 合并k个链表
class Solution {
public:struct Status {int val;ListNode *ptr;bool operator < (const Status &rhs) const {return val > rhs.val;}};priority_queue <Status> q;ListNode* mergeKLists(vector<ListNode*>& lists) {for (auto node: lists) {if (node) q.push({node->val, node});}ListNode head, *tail = &head;while (!q.empty()) {auto f = q.top(); q.pop();tail->next = f.ptr; tail = tail->next;if (f.ptr->next) q.push({f.ptr->next->val, f.ptr->next});}return head.next;}
};
4. 将数组和减半的最小次数
2208. 将数组和减半的最少操作次数 - 力扣（LeetCode）
class Solution {
public:int halveArray(vector<int>& nums) {priority_queue<long> q;long sum=0;for(int i=0;i<nums.size();i++){long tmp=(long)nums[i]<<20;q.push(tmp);sum+=tmp;}long sum2=sum>>1;int con=0;while(sum2>0){long m=q.top();q.pop();sum2-=m/2;q.push(m/2);con++;} return con;}
};5. 交换数字
面试题 16.01. 交换数字 - 力扣（LeetCode）
class Solution {
public:vector<int> swapNumbers(vector<int>& numbers) {numbers[0]=numbers[0]^numbers[1];numbers[1]=numbers[0]^numbers[1];numbers[0]=numbers[0]^numbers[1];;return numbers;}
};6. 找到所有数组中消失的数字
448. 找到所有数组中消失的数字 - 力扣（LeetCode）
class Solution {
public:vector<int> findDisappearedNumbers(vector<int>& nums) {int n = nums.size();for (auto& num : nums) {int x = (num - 1) % n;if(nums[x]<=n)nums[x] += n;}vector<int> ret;for (int i = 0; i < n; i++) {if (nums[i] <= n) {ret.push_back(i + 1);}}return ret;}
};7. 只出现一次的数字
LCR 004. 只出现一次的数字 II - 力扣（LeetCode）
class Solution {
public:int singleNumber(vector<int>& nums) {vector<int> tmp(32,0);for(auto& m: nums){for(int i=0;i<tmp.size();i++){tmp[i]+=(m>>i)&1;}}int res=0;for(int i=0;i<tmp.size();i++){if(tmp[i]%3!=0){res|=1<<i;}}return res;}
};