Maximum Subarray

经典的求最大连续子数组和，以下有两种方法。

方法一：动规，复杂度O（n）

一遍扫描，R[i]表示以第i个元素结尾的最大连续子数组和，那么R[i+1] = max(A[i+1], R[i]+A[i+1])。代码如下：

    int maxSubArray(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int i;
        int max = A[0];
        int tmp = A[0];
        for(i = 1; i < n; i++){
            if(tmp <= 0){
                tmp = A[i];
            }
            else{
                tmp += A[i];
            }
            if(tmp > max)
                max = tmp;
        }
        return max;
    }

方法二：分治的方法，把数组从中间分开，那么最大和有三种情况：全在左边，全在右边，两边都有。复杂度为O（nlgn）。

int maxSubArray(int A[], int left, int right){
        if(left == right){
            return A[left];
        }
        int middle = (left + right)/2;
        int maxLeft = maxSubArray(A, left, middle);
        int maxRight = maxSubArray(A, middle+1, right);
        int maxToLeftBorder = A[middle];
        int i,tmp=0;
        for(i = middle; i >= 0; i--){
            tmp += A[i];
            if(tmp > maxToLeftBorder)
                maxToLeftBorder = tmp;
        }
        int maxToRightBorder = A[middle + 1];
        tmp = 0;
        for(i = middle+1; i < right; i++){
            tmp += A[i];
            if(tmp > maxToRightBorder)
                maxToRightBorder = tmp;
        }
        int max1 = maxLeft>maxRight?maxLeft:maxRight;
        int max2 = maxToLeftBorder+maxToRightBorder;
        return max1>max2?max1:max2;
    }
    int maxSubArray(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return maxSubArray(A, 0, n-1);
    }