1、题目
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
, or0 <= 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
2、代码
class Solution {
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
if (A == null || A.length == 0){
return 0;
}
int[] mediateMResult = new int[A.length - M + 1];
for (int i = 0; i < M; i++){
mediateMResult[0] += A[i];
}
for (int i = 1; i < mediateMResult.length; i++){
mediateMResult[i] = mediateMResult[i - 1] - A[i - 1] + A[i + M - 1];
}
int[] mediateLResult = new int[A.length - L + 1];
for(int i = 0; i < L;i++){
mediateLResult[0] += A[i];
}
for (int i = 1; i < mediateLResult.length; i++){
mediateLResult[i] = mediateLResult[i - 1] - A[i - 1] + A[i + L - 1];
}
int result = 0;
for (int i = 0; i < mediateMResult.length; i++){
int firstValue = mediateMResult[i];
int secondValue = Math.max(maxMediateSum(A, 0, i, mediateLResult, L), maxMediateSum(A, i + M, A.length, mediateLResult, L));
result = Math.max(result, firstValue + secondValue);
}
return result;
}
public int maxMediateSum(int[] A, int begin, int end, int[] mediateResult, int L){
if (end - begin < L){
return -1;
}
int result = mediateResult[begin];
for (int i = begin + 1; i <= end - L; i ++){
result = Math.max(result, mediateResult[i]);
}
return result;
}
}
优化版:
class Solution {
public int maxSumTwoNoOverlap(int[] A, int L, int M) {
if (A == null || A.length == 0){
return 0;
}
for(int i = 1; i < A.length; i++){
A[i] += A[i - 1];
}
int Lmax = A[L -1];
int Mmax = A[M - 1];
int res = A[L + M - 1];
for (int i = L + M ; i < A.length; i++){
Lmax = Math.max(Lmax, A[i - M] - A[i - M - L]);
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;
}
}