最大子段和 - 分治递归算法C++实现

最大字段和有很多种解法，分治递归也有其优点，这里实现一下。
时间复杂度 ：n log(n)
空间复杂度O(1)没有占用额外空间

#include <iostream>
#include <math.h>
using namespace std;
int bestI = -1, bestJ = -1,bestSum = -0x3f3f3f3f;
int* a = new int[100];
int maxSum(int* a, int left, int right) {
    
    
	if (left == right) {
    
     //终止条件，更新最佳范围，返回最大值
		if (a[left] > bestSum) {
    
    
			bestSum = a[left];
			bestI = bestJ = left;
		}
		return a[left];
	}
	else {
    
    
		int center = (left + right) / 2;
		int maxLeft = maxSum(a, left, center);
		int maxRight = maxSum(a, center + 1, right);
		
		//找左区间连续最大的
		int temporaryBestI = center, temporaryBestJ = center+1;
		int leftSum = a[center];
		int temporarySum = a[center];
		for (int i = center-1; i >= left; i--) {
    
    
			temporarySum += a[i];
			if (temporarySum > leftSum) {
    
    
				leftSum = temporarySum;
				temporaryBestI = i;
			}
		}

		//找右区间连续最大的
		int rightSum = a[center+1];
		temporarySum = a[center+1];
		for (int i = center + 2; i <= right; i++) {
    
    
			temporarySum += a[i];
			if (temporarySum > rightSum) {
    
    
				rightSum = temporarySum;
				temporaryBestJ = i;
			}
		}

		int maxSum = max(maxLeft, max(maxRight, leftSum + rightSum));
		if (maxSum > bestSum) {
    
    
			bestSum = maxSum;
			if (maxSum == leftSum + rightSum) {
    
    
				bestI = temporaryBestI;
				bestJ = temporaryBestJ;
			}
		}
		return maxSum;
	}
}

int main() {
    
    
	int i = 1;
	while (cin >> a[i]) {
    
    
		i++;
	}
	cout << maxSum(a,1,i-1) << endl;
	cout << bestI << endl << bestJ;
	return 0;
}