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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.
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.
思路：遍历每一个分割线，分别求分割线左边的最大子数组和分割线右边的最大子数组。为了实现O(n) 的时间复杂度，可以创建left[] 、right[] 数组，其中left[i] 表示从0 ~ i 的数组的子数组最大值，right[j] 表示nums.length - 1 ~ j 的数组的子数组最大值。
1 public class Solution { 2 /** 3 * @param nums: A list of integers 4 * @return: An integer denotes the sum of max two non-overlapping subarrays 5 */ 6 public int maxTwoSubArrays(ArrayList<Integer> nums) { 7 if (nums == null || nums.size() == 0) { 8 return 0; 9 } 10 int[] left = new int[nums.size()]; 11 int[] right = new int[nums.size()]; 12 int sum = 0, max = Integer.MIN_VALUE; 13 for (int i = 0; i < nums.size(); i++) { 14 sum = sum + nums.get(i); 15 max = Math.max(max, sum); 16 sum = Math.max(sum, 0); 17 left[i] = max; 18 } 19 sum = 0; 20 max = Integer.MIN_VALUE; 21 for (int i = nums.size() - 1; i >= 0; i--) { 22 sum = sum + nums.get(i); 23 max = Math.max(max, sum); 24 sum = Math.max(sum, 0); 25 right[i] = max; 26 } 27 max = Integer.MIN_VALUE; 28 for (int i = 0; i < nums.size() - 1; i++) { 29 max = Math.max(left[i] + right[i + 1], max); 30 } 31 return max; 32 } 33 }
1 public class Solution { 2 /** 3 * @param nums: A list of integers 4 * @return: An integer denotes the sum of max two non-overlapping subarrays 5 */ 6 public int maxTwoSubArrays(ArrayList<Integer> nums) { 7 // write your code 8 int size = nums.size(); 9 int[] left = new int[size]; 10 int[] right = new int[size]; 11 int sum = 0; 12 int minSum = 0; 13 int max = Integer.MIN_VALUE; 14 for(int i = 0; i < size; i++){ 15 sum += nums.get(i); 16 max = Math.max(max, sum - minSum); 17 minSum = Math.min(sum, minSum); 18 left[i] = max; 19 } 20 sum = 0; 21 minSum = 0; 22 max = Integer.MIN_VALUE; 23 for(int i = size - 1; i >= 0; i--){ 24 sum += nums.get(i); 25 max = Math.max(max, sum - minSum); 26 minSum = Math.min(sum, minSum); 27 right[i] = max; 28 } 29 max = Integer.MIN_VALUE; 30 for(int i = 0; i < size - 1; i++){ 31 max = Math.max(max, left[i] + right[i + 1]); 32 } 33 return max; 34 } 35 }
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原文地址：http://www.cnblogs.com/FLAGyuri/p/5347267.html
