最大子陣列問題求解的是給定一個數組a[0...n-1]，求出它的一個子陣列使得其所有元素的和加起來最大

如果使用暴力解法即列舉所有的子陣列，則時間複雜度為O（n^2）

採用分治法，對一段陣列a[low.....high],求它的最大子陣列，mid = （low+high）/ 2

那麼，a[low..high]的子陣列有可能有三種分佈的情況，假設子陣列的上下屆為i,j

1)  在a的左部分，即  low < =  i <= j < = mid

2) 在a的右部分，即    mid + 1 <= i  <=  j <= high

3) 跨越mid，即  low <= i <= mid <= j  <= high

下面給出基於此思想的分治演算法

注：要對一個數組a[0...n-1]求其最大子陣列和，呼叫的是find_max_subarray(a,0,n-1)

#include <iostream>
#include <limits.h>

using namespace std;

struct temple {
    int left;
    int right;
    int sum;
};

temple max_crossing_subarr ( int* a, int low, int mid, int high) {

    int left_sum = INT_MIN;
    int right_sum = INT_MIN;
    int sum;
    int max_left = low;
    int max_right = high;

    //找左邊的
    sum = 0;
    for ( int i = mid; i >= low; --i ) {
        sum += a[i];
        if ( sum > left_sum ) {
            left_sum = sum;
            max_left = i;
        }
    }

    //找右邊的
    sum = 0;
    for ( int i = mid+1; i <= high; ++i ) {
        sum += a[i];
        if ( sum > right_sum ) {
            right_sum = sum;
            max_right = i;
        }
    }

    temple t = { max_left, max_right, left_sum + right_sum };
    return t;
}

temple find_max_subarray ( int*a, int low, int high ) {

    if ( high == low ) {
        temple t;
        t.left = low;
        t.right = high;
        t.sum = a[low];
        return t;
    }
    else {

        int mid = ( high + low ) / 2;
        temple left_temple = find_max_subarray(a,low,mid);
        temple right_temple = find_max_subarray(a,mid+1,high);
        temple cross_temple = max_crossing_subarr(a,low,mid,high);

        if ( left_temple.sum >= right_temple.sum
            && left_temple.sum >= cross_temple.sum ) {
                return left_temple;
        }
        else if ( right_temple.sum >= left_temple.sum
            && right_temple.sum >= cross_temple.sum ) {
                return right_temple;
        }
        else {
            return cross_temple;
        }

    }

}


int main()
{
//    int a[] = { 13, -3, -25, 20, -3, 16, -23, 18, 20, -7, 12
//            -5, 22, 15, -4, 7};

    int a[] = {-5,1,2,5,-3,7,-10};
    temple t = find_max_subarray(a,0,sizeof(a)/sizeof(int)-1);

    cout << t.sum  << endl;//結果12

    return 0;
}