Eric Chen Mock Interview 

 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}

 

posted @ 2016-06-27 05:42  bambu  阅读(162)  评论(0)    收藏  举报