# 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<Integer> 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<Integer> 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<Integer> 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).