控制模型执行 | AnyLogic帮助

目录

排序数组（归并排序）

题解：

交易逆序对的总数

题解：

计算右侧小于当前元素的个数

题解：

翻转对

题解： 

排序数组（归并排序）

912. 排序数组 - 力扣（LeetCode）https://leetcode.cn/problems/sort-an-array/description/

题解：

class Solution {
public:
    vector<int> tmp;
    vector<int> sortArray(vector<int>& nums) {
        tmp.resize(nums.size());
        mergeSort(nums,0,nums.size()-1);//左闭右闭
        return nums;
    }
    void mergeSort(vector<int>& nums,int left,int right)
    {
        if(left>=right) return;
        int mid=left+(right-left)/2;
        mergeSort(nums,left,mid);   mergeSort(nums,mid+1,right);
        int cur1=left,cur2=mid+1,i=0;
        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++];

        for(int j=left,i=0;j<=right;j++) nums[j]=tmp[i++];   
    }

};

交易逆序对的总数

LCR 170. 交易逆序对的总数 - 力扣（LeetCode）https://leetcode.cn/problems/shu-zu-zhong-de-ni-xu-dui-lcof/description/

题解：

class Solution {
public:
    vector<int> tmp;
    int reversePairs(vector<int>& record) {
        tmp.resize(record.size());
        return mergeSort(record,0,record.size()-1);//左闭右闭
    }
    int mergeSort(vector<int>& record,int left,int right)
    {
        if(left>=right) return 0;
        int ret=0,mid=left+(right-left)/2;
        ret+=mergeSort(record,left,mid);
        ret+=mergeSort(record,mid+1,right);

        //边合并边计数,排升序
        int cur1=left,cur2=mid+1,i=0;
        while(cur1<=mid && cur2<=right)
        {
            if(record[cur1]<=record[cur2])   
                tmp[i++]=record[cur1++];
            else
            {
                ret+=(mid-cur1+1);  tmp[i++]=record[cur2++];
            }
        }
        while(cur1<=mid)   tmp[i++]=record[cur1++];
        while(cur2<=right)  tmp[i++]=record[cur2++];

        for(int j=left,i=0;j<=right;j++)    record[j]=tmp[i++];
        return ret;
    }
};

计算右侧小于当前元素的个数

315. 计算右侧小于当前元素的个数 - 力扣（LeetCode）https://leetcode.cn/problems/count-of-smaller-numbers-after-self/description/

题解：

class Solution {
public:
    vector<int> ret;
    vector<int> tmpIndex;//临时数组的下标
    vector<int> index;//原数组元素的下标
    vector<int> tmpNums;//临时数组
    vector<int> countSmaller(vector<int>& nums) {
        int n=nums.size();
        tmpIndex.resize(n);   tmpNums.resize(n);
        index.resize(n);  ret.resize(n);
        for(int i=0;i<n;i++)  index[i]=i;//初始化

        mergeSort(nums,0,n-1);//左闭右闭
        return ret;
    }
    void mergeSort(vector<int>& nums,int left,int right)
    {
        if(left>=right) return;
        int mid=left+(right-left)/2;
        mergeSort(nums,left,mid);   mergeSort(nums,mid+1,right);
        int cur1=left,cur2=mid+1,i=0;
        while(cur1<=mid && cur2<=right)//排降序
        {
            if(nums[cur1]<=nums[cur2])//右侧比当前大，只排序
            {
                tmpNums[i]=nums[cur2];
                tmpIndex[i++]=index[cur2++];
            }
            else//右侧比当前大，计数且排序
            {
                ret[index[cur1]]+=(right-cur2+1);
                tmpNums[i]=nums[cur1];
                tmpIndex[i++]=index[cur1++];
            }
        }
        while(cur1<=mid)   
        {
            tmpNums[i]=nums[cur1];
            tmpIndex[i++]=index[cur1++];
        }
        while(cur2<=right)
        {
            tmpNums[i]=nums[cur2];
            tmpIndex[i++]=index[cur2++];
        }
        //填回原数组
        for(int j=left,i=0;j<=right;j++)    
        {
            nums[j]=tmpNums[i];   index[j]=tmpIndex[i++];
        }
    }
};

翻转对

493. 翻转对 - 力扣（LeetCode）https://leetcode.cn/problems/reverse-pairs/description/

题解： 

class Solution {
public:
    vector<int> tmp;
    int reversePairs(vector<int>& nums) {
        int n=nums.size();  tmp.resize(n);
        return mergeSort(nums,0,n-1);//左闭右闭
    }
    int mergeSort(vector<int>& nums,int left,int right)
    {
        if(left>=right) return 0;
        int ret=0,mid=left+(right-left)/2;
        ret+=mergeSort(nums,left,mid);
        ret+=mergeSort(nums,mid+1,right);
        int cur1=left,cur2=mid+1,i=0;
        //找翻转对
        while(cur1<=mid)
        {
            while(cur2<=right && nums[cur2]>=nums[cur1]/2.0)    cur2++;
            if(cur2>right)  break;//没有必要遍历了
            ret+=(right-cur2+1);    cur1++;
        }
        //合并,排降序
        cur1=left,cur2=mid+1,i=0;
        while(cur1<=mid && cur2<=right)
            tmp[i++]=(nums[cur1]>nums[cur2]?nums[cur1++]:nums[cur2++]);
        while(cur1<=mid)    tmp[i++]=nums[cur1++];
        while(cur2<=right)  tmp[i++]=nums[cur2++];
        for(int j=left,i=0;j<=right;j++,i++)    nums[j]=tmp[i];
        return ret;
    }
};