leetcode算法之分治-归并

1.排序数组

排序数组
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//分治-归并
class Solution {int tmp[50010];
public:vector<int> sortArray(vector<int>& nums) {mergeSort(nums,0,nums.size()-1);return nums;}void mergeSort(vector<int>& nums,int left,int right){if(left>=right) return;//1.选择中间点划分区间int mid = (left+right)>>1;//[left,mid][mid+1,right]//2.将左右两区间排序mergeSort(nums,left,mid);mergeSort(nums,mid+1,right);//3.合并两个有序数组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++];//还原for(int j=left;j<=right;j++){nums[j] = tmp[j-left];}}
};

2.数组中的逆序对

数组中的逆序对
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class Solution {int tmp[50010];
public:int reversePairs(vector<int>& nums) {return mergeSort(nums,0,nums.size()-1);}int mergeSort(vector<int>& nums,int left,int right){if(left>=right) return 0;int ret = 0;//1.选择中间元素划分区间int mid = (left+right)>>1;//[left,mid][mid+1,right]//2.计算左右区间的逆序对的个数ret += mergeSort(nums,left,mid);ret += mergeSort(nums,mid+1,right);//3.计算一左一右逆序对的个数+合并两个有序数组int cur1 = left,cur2 = mid+1,i=0;while(cur1<=mid&& cur2<=right)//升序{if(nums[cur1]<=nums[cur2]) tmp[i++] = nums[cur1++];else{ret += mid-cur1+1;tmp[i++] = nums[cur2++];}}while(cur1<=mid) tmp[i++] = nums[cur1++];while(cur2<=right) tmp[i++] = nums[cur2++];//还原for(int j=left;j<=right;j++){nums[j] = tmp[j-left];}return ret;}
};

3.计算右侧小于当前元素的个数

计算右侧小于当前元素的个数
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class Solution {vector<int> ret;vector<int> index;//记录nums当前元素的原始下标int tmpNums[500010];int tmpIndex[500010];
public:vector<int> countSmaller(vector<int>& nums) {//计算当前元素之后，有多少个比我小(降序)int n = nums.size();ret.resize(n);index.resize(n);//初始化index数组for(int i=0;i<n;i++){index[i] = i;}mergeSort(nums,0,n-1);return ret;}void mergeSort(vector<int>& nums,int left,int right){if(left>=right) return;//1.选择中间元素划分区间int mid = (left+right)>>1;//[left,mid][mid+1,right]//2.将左右区间进行排序mergeSort(nums,left,mid);mergeSort(nums,mid+1,right);//3.处理一左一右的情况int cur1 = left,cur2 = mid+1,i=0;while(cur1<=mid&& cur2<=right)//降序{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++];}}while(cur1<=mid) {tmpNums[i] = nums[cur1];tmpIndex[i++] = index[cur1++];}while(cur2<=right){tmpNums[i] = nums[cur2];tmpIndex[i++] = index[cur2++];}//还原for(int j=left;j<=right;j++){nums[j] = tmpNums[j-left];index[j] = tmpIndex[j-left];}}
};

4.翻转对

翻转对
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//计算当前元素之前，有多少元素的一半比我大--升序
class Solution {int tmp[50010];
public:int reversePairs(vector<int>& nums) {return mergeSort(nums,0,nums.size()-1);  }int mergeSort(vector<int>& nums,int left,int right){if(left>=right) return 0;int ret = 0;//1.选择中间元素划分区间int mid = (left+right)>>1;//[left,mid][mid+1,right]//2.计算左右区间翻转对的个数ret += mergeSort(nums,left,mid);ret += mergeSort(nums,mid+1,right);//3.先计算翻转对的个数int cur1 = left,cur2 = mid+1,i=0;while(cur2<=right)//升序的情况{while(cur1<=mid && nums[cur1]/2.0<=nums[cur2]) cur1++;if(cur1 > mid) break;ret += mid-cur1+1;cur2++;}//4.合并两个有序数组cur1 = left,cur2 = mid+1;while(cur1<=mid && cur2<=right)//升序{tmp[i++] = nums[cur1]<=nums[cur2]?nums[cur1++]:nums[cur2++];}while(cur1<=mid) tmp[i++] = nums[cur1++];while(cur2<=right) tmp[i++] = nums[cur2++];//还原for(int j=left;j<=right;j++){nums[j] = tmp[j-left];}return ret;}
};

//计算当前元素后面，有多少元素的两倍比我小--降序
class Solution {int tmp[50010];
public:int reversePairs(vector<int>& nums) {return mergeSort(nums,0,nums.size()-1);  }int mergeSort(vector<int>& nums,int left,int right){if(left>=right) return 0;int ret = 0;//1.选择中间元素划分区间int mid = (left+right)>>1;//[left,mid][mid+1,right]//2.计算左右区间翻转对的个数ret += mergeSort(nums,left,mid);ret += mergeSort(nums,mid+1,right);//3.先计算翻转对的个数int cur1 = left,cur2 = mid+1,i=0;while(cur1<=mid)//降序的情况{while(cur2<=right && nums[cur1]/2.0<=nums[cur2]) cur2++;if(cur2 > right) break;ret +=right-cur2+1;cur1++;}//4.合并两个有序数组cur1 = left,cur2 = mid+1;while(cur1<=mid && cur2<=right)//降序{tmp[i++] = nums[cur1]<=nums[cur2]?nums[cur2++]:nums[cur1++];}while(cur1<=mid) tmp[i++] = nums[cur1++];while(cur2<=right) tmp[i++] = nums[cur2++];//还原for(int j=left;j<=right;j++){nums[j] = tmp[j-left];}return ret;}
};