技術標籤：c++資料結構與演算法c++

裡面有句抖機靈

		int mid = left + ((right - left) >> 1);

可以理解為

mid=(L+R)/2

但是這是一種不安全的寫法，因為可能(R+L)溢位
安全的寫法可以表示為:

mid=L+(R-L)/2;

而除以2就可以寫成“>>1”：二進位制右移一位
位運算的時間比常數運算要好

#include <iostream>
#include <vector>
#include <ctime>
using namespace std;

class CSmallSum
 
 {
public:
	int smallSum(vector<int>& arr) {
		if (arr.size() < 2) {
			return 0;
		}
		return mergeSortProcess(arr, 0, arr.size() - 1);
	}

private:
	int mergeSortProcess(vector<int>& arr, int left, int right) {
		if (left == right) {
			return 0;
		}
		int mid = left + ((right -
 
 left) >> 1);
		return mergeSortProcess(arr, left, mid)
			+ mergeSortProcess(arr, mid + 1, right)
			+ merge(arr, left, mid, right);
	}

	int merge(vector<int>& arr, int left, int mid, int right) {
		int tmpL = left;
		int tmpR = mid + 1;
		int res = 0;
		vector<int> help;
		while
 
 (tmpL <= mid && tmpR <= right) {
			// 生成小和的關鍵一步
			res += arr.at(tmpL) < arr.at(tmpR) ? arr.at(tmpL) * (right - tmpR + 1) : 0;
			help.push_back(arr.at(tmpL) < arr.at(tmpR) ? arr.at(tmpL++) : arr.at(tmpR++));
		}
		while (tmpL <= mid) {
			help.push_back(arr.at(tmpL++));
		}
		while (tmpR <= right) {
			help.push_back(arr.at(tmpR++));
		}
		for (int i = 0; i < help.size(); i++) {
			arr.at(left + i) = help.at(i);
		}
		return res;
	}
};

int main() {
	vector<int> arr{ 1, 3, 4, 2, 5 };
	int res = CSmallSum().smallSum(arr);
	cout << res << endl;

	system("pause");
	return 0;
}