Maximum Subarray II
最后更新于：2022-04-02 01:13:29
# Maximum Subarray II
### Source
- lintcode: [(42) Maximum Subarray II](http://www.lintcode.com/en/problem/maximum-subarray-ii/)
~~~
Given an array of integers,
find two non-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.
Example
For given [1, 3, -1, 2, -1, 2],
the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2],
they both have the largest sum 7.
Note
The subarray should contain at least one number
Challenge
Can you do it in time complexity O(n) ?
~~~
### 题解
严格来讲这道题这道题也可以不用动规来做，这里还是采用经典的动规解法。[Maximum Subarray](http://algorithm.yuanbin.me/zh-cn/dynamic_programming/maximum_subarray.html) 中要求的是数组中最大子数组和，这里是求不相重叠的两个子数组和的和最大值，做过买卖股票系列的题的话这道题就非常容易了，既然我们已经求出了单一子数组的最大和，那么我们使用隔板法将数组一分为二，分别求这两段的最大子数组和，求相加后的最大值即为最终结果。隔板前半部分的最大子数组和很容易求得，但是后半部分难道需要将索引从0开始依次计算吗？NO!!! 我们可以采用从后往前的方式进行遍历，这样时间复杂度就大大降低了。
### Java
~~~
public class Solution {
/**
* @param nums: A list of integers
* @return: An integer denotes the sum of max two non-overlapping subarrays
*/
public int maxTwoSubArrays(ArrayList nums) {
// -1 is not proper for illegal input
if (nums == null || nums.isEmpty()) return -1;
int size = nums.size();
// get max sub array forward
int[] maxSubArrayF = new int[size];
forwardTraversal(nums, maxSubArrayF);
// get max sub array backward
int[] maxSubArrayB = new int[size];
backwardTraversal(nums, maxSubArrayB);
// get maximum subarray by iteration
int maxTwoSub = Integer.MIN_VALUE;
for (int i = 0; i < size - 1; i++) {
// non-overlapping
maxTwoSub = Math.max(maxTwoSub, maxSubArrayF[i] + maxSubArrayB[i + 1]);
}
return maxTwoSub;
}
private void forwardTraversal(List nums, int[] maxSubArray) {
int sum = 0, minSum = 0, maxSub = Integer.MIN_VALUE;
int size = nums.size();
for (int i = 0; i < size; i++) {
minSum = Math.min(minSum, sum);
sum += nums.get(i);
maxSub = Math.max(maxSub, sum - minSum);
maxSubArray[i] = maxSub;
}
}
private void backwardTraversal(List nums, int[] maxSubArray) {
int sum = 0, minSum = 0, maxSub = Integer.MIN_VALUE;
int size = nums.size();
for (int i = size - 1; i >= 0; i--) {
minSum = Math.min(minSum, sum);
sum += nums.get(i);
maxSub = Math.max(maxSub, sum - minSum);
maxSubArray[i] = maxSub;
}
}
}
~~~
### 源码分析
前向搜索和逆向搜索我们使用私有方法实现，可读性更高。注意是求非重叠子数组和，故求`maxTwoSub`时i 的范围为`0, size - 2`, 前向数组索引为 i, 后向索引为 i + 1.
### 复杂度分析
前向和后向搜索求得最大子数组和，时间复杂度 O(2n)=O(n)O(2n)=O(n)O(2n)=O(n), 空间复杂度 O(n)O(n)O(n). 遍历子数组和的数组求最终两个子数组和的最大值，时间复杂度 O(n)O(n)O(n). 故总的时间复杂度为 O(n)O(n)O(n), 空间复杂度 O(n)O(n)O(n).
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