【优选算法】分治

一：颜色分类

class Solution {
public:void sortColors(vector<int>& nums) {// 三指针法int n = nums.size();int left = -1, right = n, i = 0;while(i < right){if(nums[i] == 0) swap(nums[++left], nums[i++]);else if(nums[i] == 2) swap(nums[--right], nums[i]);else i++;}}
};

二：排序数组

class Solution {
public:vector<int> sortArray(vector<int>& nums) {srand(time(nullptr));qsort(nums, 0, nums.size()-1);return nums;}void qsort(vector<int>& nums, int left, int right){// 快速排序——三区间[<key][=key][>key]if(left >= right) return;// 选择随机数 Keyint key = GetRand(nums, left, right);// 三指针排法int i = left, l = left-1, r = right+1;while(i < r){if(nums[i] < key) swap(nums[++l], nums[i++]);else if(nums[i] > key) swap(nums[--r], nums[i]);else i++;} // // 开始往下快排// [left, l][l+1, r-1][r, right]qsort(nums, left, l);qsort(nums, r, right);}int GetRand(vector<int>& nums, int left, int right){int r = rand();return nums[r%(right-left+1)+left];}
};

三：数组中的第K个最大元素

 

class Solution {int GetRand(vector<int>& nums, int left, int right){int r = rand();return nums[r%(right-left+1) + left];        }void quickSort(vector<int>& nums, int left, int right){if(left > right)return;// 随机数选keyint key = GetRand(nums, left, right);// 三指针操作法// [<key][=key][>key]int l = left-1, i = left, r = right+1;while(i < r){if(nums[i] < key)swap(nums[++l], nums[i++]);else if(nums[i] > key)swap(nums[--r], nums[i]);elsei++;}// [left, l][l+1, r-1][r, right]quickSort(nums, left, l);quickSort(nums, r, right);}
public:int findKthLargest(vector<int>& nums, int k) {srand(time(0));quickSort(nums, 0, nums.size()-1);return nums[nums.size() - k];}
};

四：库存管理III

class Solution {
private:int GetRand(vector<int>& nums, int left, int right){int r = rand();return nums[r%(right-left+1) + left];}void quickSort(vector<int>& nums, int left, int right){if(left > right)return;// 获取随机数int key = GetRand(nums, left, right);// 三指针法int i = left, l = left-1, r = right+1;while(i < r){if(nums[i] > key)swap(nums[--r], nums[i]);else if(nums[i] < key)swap(nums[++l], nums[i++]);elsei++;}// []][][]quickSort(nums, left, l);quickSort(nums, r, right);}
public:vector<int> inventoryManagement(vector<int>& stock, int cnt) {srand(time(0));quickSort(stock, 0, stock.size()-1);vector<int> ret(stock.begin(), stock.begin() + cnt);return ret;}
};

五：交易逆序对的总数

 

class Solution {int tmp[50010];
private:int mergeSort(vector<int>& nums, int left, int right){if(left >= right)return 0;// 找中间点int mid = (left+right)>>1;int ret = 0;    // 要返回的结果// [left, mid][mid+1, right]// 2. 左边的个数 + 排序 +  右边的个数 + 排序ret += mergeSort(nums, left, mid);ret += mergeSort(nums, mid + 1, right);// 3. 一左一右的个数int cur1 = left, cur2 = mid + 1, i = 0;while(cur1 <= mid && cur2 <= right){if(nums[cur1] <= nums[cur2])tmp[i++] = nums[cur1++];else{ret += (mid - cur1 + 1);tmp[i++] = nums[cur2++];}}// 4. 处理一下排序while(cur1 <= mid) tmp[i++] = nums[cur1++];while(cur2 <= right) tmp[i++] = nums[cur2++];for(int i = left; i <= right; i++)nums[i] = tmp[i-left];return ret;}
public:int reversePairs(vector<int>& record) {return mergeSort(record, 0, record.size()-1);}
};

六：计算右侧小于当前元素的个数

 

class Solution {vector<int> ret;vector<int> index;  // 记录 nums 下当前元素的原始下标int tmpNums[50010];int tmpIndex[50010];    
private:void mergeSort(vector<int>& nums, int left, int right){if(left >= right)return ;int mid = (left + right) >> 1;// [left, mid][mid+1, right]// 1. 先处理 左右两部分mergeSort(nums, left, mid);mergeSort(nums, mid + 1, right);// 2. 处理一左一右的情况int cur1 = left, cur2 = mid + 1, i = 0;while(cur1 <= mid && cur2 <= right){if(nums[cur1] <= nums[cur2]){tmpNums[i] = nums[cur2];tmpIndex[i++] = index[cur2++];}else{ret[index[cur1]] = right-cur2 + 1;tmpNums[i] = nums[cur1];tmpIndex[i++] = index[cur1++];}}// 3. 处理剩下的排序while(cur1 <= mid){tmpNums[i] = nums[cur1];tmpIndex[i++] = index[cur1++];}while(cur2 <= right){tmpNums[i] = nums[cur2++];tmpIndex[i++] = index[cur2++];}for(int i = left; i <= right; i++){nums[i] = tmpNums[i-left];index[i] = tmpIndex[i-left];}}
public:vector<int> countSmaller(vector<int>& nums) {int n = nums.size();ret.resize(n);// 初始化 index 按钮for(int i = 0; i < n; i++){index[i] = i;}mergeSort(nums, 0, n-1);return ret;}
};

七：翻转对

 

class Solution {int tmp[50010];int mergeSort(vector<int>& nums, int left, int right){if(left >= right)return 0;//1int mid = (left + right)>>1;// [left, mid][mid+1, right]int ret = 0;ret += mergeSort(nums, left, mid);ret += mergeSort(nums, mid + 1, right);// 2. int cur1 = left, cur2 = mid +1, i = left;while(cur1 <= mid)  // 降序的情况{while(cur2 <= right && nums[cur2] >= nums[cur1]/2.0)cur2++;if(cur2 > right)break;ret += right - cur2 + 1;cur1++;}// 4. cur1 = left, cur2 = mid+1;while(cur1 <= mid && cur2 <= right)tmp[i++] = nums[cur1]<nums[cur2]? nums[cur2++] : nums[cur1++];while(cur1 <= mid)tmp[i++] = nums[cur1++];while(cur2 <= right)tmp[i++] = nums[cur2++];// 5for(int i = left; i <= right; i++)nums[i] = tmp[i];// 6return ret;}
public:int reversePairs(vector<int>& nums) {return mergeSort(nums, 0, nums.size()-1);}
};