FB面经prepare: 3 Window Max Sum

Given a num array, find a window of size k, that has a maximum sum of the k entries.
follow-up: Find three non-overlap windows of size k, that together has a maximum sum of the 3k entries, time complexity O(n^2)

Follow Up:

这个其实可以优化到O(n)时间。
建从左端到每个下标的最大window数组，再建从右端到每个下标的最大window数组。
再从左往右走一遍所有的size k window，将其和与上一步建的两个数组一起加起来。遍历一遍取最大值即可。

 package fbOnsite;

 public class ThreeWindow {
     public static int find(int[] arr, int k) {
         int res = Integer.MIN_VALUE;
         int len = arr.length;
         int[] left = new int[len];

         int lsum = 0;
         for (int i=0; i<len; i++) {
             lsum += arr[i];
             if (i >= k) lsum -= arr[i-k];
             if (i >= k-1) left[i] = Math.max(lsum, i>=1? left[i-1] : 0);//find the window end at i with max sum
         }

         int[] right = new int[len];
         int rsum = 0;
         for (int j=len-1; j>=0; j--) {
             rsum += arr[j];
             if (j < len-k) rsum -= arr[j+k];
             if (j <= len-k) right[j] = Math.max(rsum, j<len-1? right[j+1] : 0);
         }

         int midSum = 0;
         for (int t=k; t<=len-k-1; t++) {
             midSum += arr[t];
             if (t >= k + k) midSum -= arr[t-k];
             if (t >= k + k - 1) { // for each size k window in the middle with sum as midSum
                 //find midSum + left[t-k] + right[t+1] that gives the max
                 res = Math.max(res, midSum + left[t-k] + right[t+1]);
             }
         }
         return res;
     }

     /**
      * @param args
      */
     public static void main(String[] args) {
         // TODO Auto-generated method stub
         int[] arr = new int[]{2,1,1,-1,3,6,-2,-4,1,2,-1,2,1,-1,2};
         int[] arr1 = new int[]{1,2,2,-1,1,2,1,3,5};
         int res = find(arr, 3);
         System.out.println(res);
     }

 }