FB面经Prepare: Dot Product

Conduct Dot Product of two large Vectors

1. two pointers

2. hashmap

3. 如果没有额外空间，如果一个很大，一个很小，适合scan小的，并且在大的里面做binary search

 1 package fb;
 2 
 3 public class DotProduct {
 4     
 5     public int dotPro(int[][] v1, int[][] v2) {
 6         int[][] shortV;
 7         int[][] longV;
 8         if (v1.length < v2.length) {
 9             shortV = v1;
10             longV = v2;
11         }
12         else {
13             shortV = v2;
14             longV = v1;
15         }
16         
17         int res = 0;
18         for (int i=0; i<shortV.length; i++) {
19             int shortIndex = shortV[i][0];
20             int shortValue = shortV[i][1];
21             int longSeq = binarySearch(longV, shortIndex);
22             if (longSeq >= 0) {
23                 res += shortValue * longV[longSeq][1];
24             }
25         }
26         return res;
27     }
28     
29     public int binarySearch(int[][] arr, int target) {
30         int l=0, r=arr.length-1;
31         while (l <= r) {
32             int m = (l+r)/2;
33             if (arr[m][0] == target) return m;
34             else if (arr[m][0] < target) l = m + 1;
35             else r = m - 1;
36         }
37         return -1;
38     }
39     
40 
41     /**
42      * @param args
43      */
44     public static void main(String[] args) {
45         // TODO Auto-generated method stub
46         DotProduct sol = new DotProduct();
47         int[][] v2 = new int[][]{{0,2},{1,3},{5,2},{7,1},{10,1}};
48         int[][] v1 = new int[][]{{1,6},{7,2}};
49         int res = sol.dotPro(v1, v2);
50         System.out.println(res);
51     }
52 
53 }

posted @ 2017-02-15 06:13 neverlandly 阅读(1612) 评论(0) 编辑收藏举报

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