求第K大的数—利用快速排序的思想

class Solution {
public:int partition(vector<int>& num, int l, int r){swap(num[l],num[l+rand()%(r-l+1)]);int x = num[l];int i=l+1, j=r;while(i<=j){while(i<=r&&num[i]<=x){i++;}while(num[j]>x){j--;}if(i>=j) break;swap(num[i],num[j]);}swap(num[l],num[j]);return j;}int find(vector<int>& num, int k){int l=0,r=num.size()-1;while(l<=r){int pos = partition(num,l,r);if(pos==k-1) return num[pos];else if(pos>k-1){r = pos-1;}else{l = pos+1;}}return -1;}int findKthLargest(vector<int>& nums, int k) {return find(nums,nums.size()-k+1);}
};

求逆序对的数量-归并排序
思路是：一个数组逆序对的数量等于左部分+右部分+横跨中间的逆序对数量

#include<iostream>
using namespace std;int merge_sort(int *num, int *tmp, int l, int r){if(l==r) return 0;int mid = (l+r)/2;int res = merge_sort(num,tmp,l,mid) + merge_sort(num,tmp,mid+1,r);for(int i=l;i<=r;i++) tmp[i]=num[i];int i = l, j = mid+1;for(int k=l;k<=r;k++){if(i==mid+1){num[k]=tmp[j++];}else if(j==r+1){num[k]=tmp[i++];}else if(num[i]>num[j]){num[k]=tmp[j++];res += mid-i+1;}else{num[k]=tmp[i++];}}return res;
}
int main(){int n;scanf("%d",&n);int num[n];int tmp[n];for(int i=0;i<n;i++) scanf("%d",&num[i]);int result = merge_sort(num,tmp,0,n-1);cout<<result;//for(int i=0;i<n;i++) printf("%d ",num[i]);return 0;
}