1031. Maximum Sum of Two Non-Overlapping Subarrays

问题：

给定数组，求两个固定长度M和L的子数组（不相交）之和最大为多少

Example 1:
Input: A = [0,6,5,2,2,5,1,9,4], L = 1, M = 2
Output: 20
Explanation: One choice of subarrays is [9] with length 1, and [6,5] with length 2.

Example 2:
Input: A = [3,8,1,3,2,1,8,9,0], L = 3, M = 2
Output: 29
Explanation: One choice of subarrays is [3,8,1] with length 3, and [8,9] with length 2.

Example 3:
Input: A = [2,1,5,6,0,9,5,0,3,8], L = 4, M = 3
Output: 31
Explanation: One choice of subarrays is [5,6,0,9] with length 4, and [3,8] with length 3. 

Note:
L >= 1
M >= 1
L + M <= A.length <= 1000
0 <= A[i] <= 1000

　　

解法：

L和M的关系可能为以下两种：

[..L..M..]

[..M..L..]

我们可以固定右边的子数组M(或L)

求左边的子数组最大和Lmax(或Mmax)

对每一组Lmax+M，求最大和，再向右移动M

同样，对每一组Mmax+L，求最大和，再向右移动L

再对这两种情况求最大值。res=max(res, max(Lmax+M, Mmax+L))

另，子数组求和，我们可利用前缀数组求和法。

sum(i~j)=presum(j)-presum(i-1)

代码参考：

class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        int Lmax=0;//L->M
        int Mmax=0;//M->L
        //sum[i-j]=presum[j]-presum[i-1]
        int res=0;
        for(int i=1; i<A.size(); i++){
            A[i]+=A[i-1];
        }
        Lmax=A[L-1];
        Mmax=A[M-1];
        res=A[M+L-1];
        for(int i=L+M; i<A.size(); i++){
            Lmax=max(Lmax, A[i-M]-A[i-M-L]);
            Mmax=max(Mmax, A[i-L]-A[i-L-M]);
            res=max(res, max(Lmax+A[i]-A[i-M], Mmax+A[i]-A[i-L]));
        }
        return res;
    }
};