[Leetcode 1031] Maximum Sum of Two Non-Overlapping Subarrays

原题说明

Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. (For clarification, the L-length subarray could occur before or after the M-length subarray.)

Formally, return the largest V for which V = (A[i] + A[i+1] + ... + A[i+L-1]) + (A[j] + A[j+1] + ... + A[j+M-1]) and either:
0 <= i < i + L - 1 < j < j + M - 1 < A.length, or
0 <= j < j + M - 1 < i < i + L - 1 < A.length.

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

解题思路

动态规划
凡是要求数组某一段的和，要想到用pre_sum，pre_sum[i]表示指数i之前左右数的和。
这样我们要求A[i]到A[j]之间所有数的和，就可以用pre_sum[j] - pre_sum[i - 1]
回到这道题，我们第一遍遍历数组A求pre_sum。

第二遍遍历，指数为i，用max_L记录指数i - M之前的最大连续L个数之和，
每次更新max_L为max(max_L, pre_sum[i - M] - pre_sum[i - L - M])
max_L + pre_sum[i] - pre_sum[i - M]表示以i结尾最后连续M个数，与之前最大的连续L个数的和。
同理max_M + pre_sum[i] - pre_sum[i - L]就表示以i结尾最后连续L个数，与之前最大的连续M个数的和。
取其中较大的与最终要返回的值res比较，并更新即可。

示例代码 (cpp)

class Solution {
public:
    int maxSumTwoNoOverlap(vector<int>& A, int L, int M) {
        if (A.size() < L + M) {
            return 0;
        }
        vector<int> pre_sum = A;
        for (int i = 1; i < A.size(); ++i) {
            pre_sum[i] += pre_sum[i - 1];
        }
        int res = pre_sum[L + M - 1], max_L = pre_sum[L - 1], max_M = pre_sum[M - 1];
        for (int i = L + M; i < A.size(); ++i) {
            max_L = max(max_L, pre_sum[i - M] - pre_sum[i - L - M]);
            max_M = max(max_M, pre_sum[i - L] - pre_sum[i - L - M]);
            res = max(res, max(max_L + pre_sum[i] - pre_sum[i - M], max_M + pre_sum[i] - pre_sum[i - L]));
        }
        return res;
    }
};

示例代码 (java)

class Solution {
    public int maxSumTwoNoOverlap(int[] A, int L, int M) {
        for (int i = 1; i < A.length; ++i)
            A[i] += A[i - 1];
        int res = A[L + M - 1], Lmax = A[L - 1], Mmax = A[M - 1];
        for (int i = L + M; i < A.length; ++i) {
            Lmax = Math.max(Lmax, A[i - M] - A[i - L - M]);
            Mmax = Math.max(Mmax, A[i - L] - A[i - L - M]);
            res = Math.max(res, Math.max(Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L]));
        }
        return res;
    }
}

示例代码 (python)

class Solution(object):
    def maxSumTwoNoOverlap(self, A, L, M):
        """
        :type A: List[int]
        :type L: int
        :type M: int
        :rtype: int
        """
        for i in xrange(1, len(A)):
            A[i] += A[i - 1]
        res, Lmax, Mmax = A[L + M - 1], A[L - 1], A[M - 1]
        for i in xrange(L + M, len(A)):
            Lmax = max(Lmax, A[i - M] - A[i - L - M])
            Mmax = max(Mmax, A[i - L] - A[i - L - M])
            res = max(res, Lmax + A[i] - A[i - M], Mmax + A[i] - A[i - L])
        return res

复杂度分析

时间复杂度: O(N)
空间复杂度: O(N)

归纳总结

我们在Youtube上更新了视频讲解，欢迎关注！