Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

題目解析：

這道題之前已經總結過了：最大連續子陣列和。但可能當時沒理解到位，導致動態規劃時又出錯誤，分不清什麼時候判斷sum。

利用動態規劃思想，sum[i]表示以A[i]結尾的最大連續子陣列和。好，那麼sum[i]什麼時候最大？sum[i] = max(sum[i-1]+A[i],A[i])。再進一步化簡的話，就是sum[i-1]是否小於0，當小於0時，sum[i] = A[i]。也就是求出的sum[i]一定包含A[i]的值。

所以在for迴圈中一開始就判斷sum[i-1]的值。又因為求sum[i]只和前一個值有關係，又可以進一步化簡，只用一個變數，而不用申請一段空間。

class Solution {
public:
    int maxSubArray(int A[], int n) {
        if(n<=0)
            return 0;
        int sum = A[0];
        int max = A[0];
        for(int i = 1;i < n;i++){
            sum += A[i];
            if(sum < 0){
                sum = 0;
            }
            if(sum > max)
                max = sum;
        }
        return max;

    }
};
 

方案二：分治法

題目中要求用分治策略，在上面連結中已經有了，要注意跨越中間點時，左右兩邊一定要加上資料。

int max_sum_array2(int arr[],int low,int high,int *left,int *right)
{
    if(low == high){    //遞迴結束點
        *left = low;
        *right = high;
        return arr[low];
    }
    int mid = (low+high)/2;
    int lleft,lright,rleft,rright,mleft,mright;
    //三種情況下分別求解。
    int lmax = max_sum_array2(arr,low,mid,&lleft,&lright);
    int rmax = max_sum_array2(arr,mid+1,high,&rleft,&rright);
    int mmax = find_max_crossing_subarray(arr,low,mid,high,&mleft,&mright);
    //返回最大值
    if(lmax > rmax && lmax > mmax){
        *left = lleft;
        *right = lright;
        return lmax;
    }else if(rmax > lmax && rmax > mmax){
        *left = rleft;
        *right = rright;
        return rmax;
    }else{
        *left = mleft;
        *right = mright;
        return mmax;
    }
}

int find_max_crossing_subarray(int arr[],int low,int mid,int high,int *left,int *right)
{
    int lsum = -100;
    int sum = 0;

    for(int i=mid;i>=low;i--){  //這裡的演算法類似一趟暴力演算法，求出最大值
        sum += arr[i];
        if(sum > lsum){
            lsum = sum;
            *left = i;
        }
    }

    int rsum = -100;
    sum = 0;
    for(int j=mid+1;j<=high;j++){
        sum += arr[j];
        if(sum > rsum){
            rsum = sum;
            *right = j;
        }
    }

    return lsum+rsum;
}