分治法求最大子序列

给定一个序列，下标为 i, i+1, i+2, ...... , j，设 mid = (i+j)/2， 则最大子序列可能出现的地方有三个，mid的左边，mid的右边，或者在中间（包括mid）。只要求出左边和右边的最大子序列（子问题），和边界上左边和右边最大子序列的和，找出三个子序列中最大的即可。

#include <iostream>
using namespace std;
/*分治法解决最大子序列问题*/
int MaxSubSum(const int a[], int left, int right);        

int main(){

    freopen("input.txt", "r", stdin);
    int a[100];
    int num;
    int length = 0;
    while(cin >> num){
        a[length++] = num;
    }

    cout << MaxSubSum(a, 0, length-1) << endl;

}

int MaxSubSum(const int a[], int left, int right){

    int maxLeftSum, maxRightSum;
    int maxLeftBorderSum = 0, maxRightBorderSum = 0;
    int leftBorderSum = 0, rightBorderSum = 0;
    int center = (left + right)/2;

    //基准情况
    if(left == right){
        if(a[left] > 0)
            return a[left];
        else
            return 0;
    }

    //递归调用子序列（进行分治）
    maxLeftSum = MaxSubSum(a, left, center);
    maxRightSum = MaxSubSum(a, center+1, right);

    //跨越中间的最长子序列和
    for(int i=center; i>=left; i--){
        leftBorderSum += a[i];
        if(leftBorderSum > maxLeftBorderSum){
            maxLeftBorderSum = leftBorderSum;
        }
    }

    for(int i=center+1; i<=right; i++){
        rightBorderSum += a[i];
        if(rightBorderSum > maxRightBorderSum){
            maxRightBorderSum = rightBorderSum;
        }
    }

    //求出这三种情况下最大的
    return max(max(maxLeftSum, maxRightSum), (maxLeftBorderSum + maxRightBorderSum)); 

}

时间复杂度为 O(NlogN)