315. Count of Smaller Numbers After Self ( Divide and Conquer)
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题目:
You are given an integer array nums and you have to return a new counts array. The counts array has the property where counts[i]
is the number of smaller elements to the right of nums[i]
.
Example:
Given nums = [5, 2, 6, 1]To the right of 5 there are 2 smaller elements (2 and 1).To the right of 2 there is only 1 smaller element (1).To the right of 6 there is 1 smaller element (1).To the right of 1 there is 0 smaller element.
Return the array [2, 1, 1, 0]
.
翻译:
您将获得一个整数数组nums,您必须返回一个新的计数数组。 计数数组具有其中count [i]是nums [i]右边较小元素的数量的属性。
例:
给定nums = [5,2,6,1]
在5的右侧有2个较小的元素(2和1)。
在右侧,只有1个较小的元素(1)。
在6的右侧有1个较小的元素(1)。
在1的右侧有0个较小的元素。
返回数组[2,1,1,0]。
解题思路:
1.另开一个arr数组赋上nums的值并排序
2.循环nums[] 二叉查找在排序后的arr中分位置即为在其右的较小个数,并在arr中删除此数;
3.循环得到的结果即为所求。
易错点:
可能存在[1,1] 重复的情况, 所以二叉查找要查找最小的位置。
代码:
#include<iostream>#include<map>#include<algorithm>#include<vector>#include<list> using namespace std;class Solution {int binaryFind(vector<int> &lis,int target) {int len = lis.size();int l = 0, r = len-1,middle = (l+r)/2;//list<int>::iterator begin = lis.begin();while(l<r) {middle = (l+r)/2;if(lis[middle] < target) l = middle+1;else r = middle;}//cout << l<<endl;return l;}public: vector<int> countSmaller(vector<int>& nums) { int len = nums.size();int *arr=new int[len];for(int i = 0; i < len;i++) arr[i] = nums[i];sort(arr,arr+len);vector<int> l(arr,arr+len); for(int i = 0;i < len;i++) { int tem = binaryFind(l,nums[i]); arr[i] = tem;l.erase(l.begin()+tem);}return vector<int>(arr,arr+len); }};int main() {Solution s;int arr[]= {5,2,6,1};vector<int> v(arr,arr+4);vector<int> tem = s.countSmaller(v);for(int i = 0; i< tem.size();i++) cout << tem[i] <<" ";}
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