1 Given an array with integers.
2 Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)| is the largest.
3 Return the largest difference.
4 1,-2,3,-1
5 1，3 A
6 -1 B
7 7
8
9 public int maxDiffSubArrays(int[] nums) {
10 // write your code here
11
12 }
13
14 Solution:
15
16 public int maxDiffSubArrays(ArrayList<Integer> nums) {
17 // write your code
18 if (nums==null || nums.size()==0) return 0;
19 int len = nums.size();
20 int[] lGlobalMax = new int[len];
21 int[] lGlobalMin = new int[len];
22 int lLocalMax = nums.get(0);
23 int lLocalMin = nums.get(0);
24 lGlobalMax[0] = lLocalMax;
25 lGlobalMin[0] = lLocalMin;
26 for (int i=1; i<len; i++) {
27 lLocalMax = Math.max(lLocalMax+nums.get(i), nums.get(i)); // 常规算 dp 的数组
28 lGlobalMax[i] = Math.max(lLocalMax, lGlobalMax[i-1]); // 将 lLocalMax 的结果变成 不以 nums[i] 结尾的数组， 方便最后的 maxHead[0, i], minEnd[i+1, n-1] 的O（n） 遍历。
29 lLocalMin = Math.min(lLocalMin+nums.get(i), nums.get(i));
30 lGlobalMin[i] = Math.min(lLocalMin, lGlobalMin[i-1]);
31 }
32 int[] rGlobalMax = new int[len];
33 int[] rGlobalMin = new int[len];
34 int rLocalMax = nums.get(len-1);
35 int rLocalMin = nums.get(len-1);
36 rGlobalMax[len-1] = rLocalMax;
37 rGlobalMin[len-1] = rLocalMin;
38 for (int i=len-2; i>=0; i--) {
39 rLocalMax = Math.max(rLocalMax+nums.get(i), nums.get(i));
40 rGlobalMax[i] = Math.max(rLocalMax, rGlobalMax[i+1]);
41 rLocalMin = Math.min(rLocalMin+nums.get(i), nums.get(i));
42 rGlobalMin[i] = Math.min(rLocalMin, rGlobalMin[i+1]);
43 }
44
45 int maxDiff = Integer.MIN_VALUE;
46 for (int i=0; i<len-1; i++) {
47 if (maxDiff < Math.abs(lGlobalMax[i]-rGlobalMin[i+1]))
48 maxDiff = Math.abs(lGlobalMax[i]-rGlobalMin[i+1]);
49
50 if (maxDiff < Math.abs(lGlobalMin[i]-rGlobalMax[i+1]))
51 maxDiff = Math.abs(lGlobalMin[i]-rGlobalMax[i+1]);
52 }
53 return maxDiff;
54}
