1570. Dot Product of Two Sparse Vectors

Given two sparse vectors, compute their dot product.

Implement class SparseVector:

  • SparseVector(nums) Initializes the object with the vector nums
  • dotProduct(vec) Compute the dot product between the instance of SparseVector and vec

A sparse vector is a vector that has mostly zero values, you should store the sparse vector efficiently and compute the dot product between two SparseVector.

Follow up: What if only one of the vectors is sparse?

 

Example 1:

Input: nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0]
Output: 8
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8

Example 2:

Input: nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2]
Output: 0
Explanation: v1 = SparseVector(nums1) , v2 = SparseVector(nums2)
v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0

Example 3:

Input: nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4]
Output: 6

 1 class SparseVector {
 2     Map<Integer, Integer> indexMap = new HashMap<>();
 3     SparseVector(int[] nums) {
 4         for (int i = 0; i < nums.length; i++)
 5             if (nums[i] != 0)
 6                 indexMap.put(i, nums[i]);
 7     }
 8 
 9     public int dotProduct(SparseVector vec) {
10         if (indexMap.size() == 0 || vec.indexMap.size() == 0) {
11             return 0;
12         }
13             
14         if (indexMap.size() > vec.indexMap.size()) {
15             return vec.dotProduct(this);
16         }
17             
18         int productSum = 0;
19         for (Map.Entry<Integer, Integer> entry : indexMap.entrySet()) {
20             int index = entry.getKey();
21             Integer vecValue = vec.indexMap.get(index);
22             if (vecValue != null) {
23                 productSum += (entry.getValue() * vecValue);
24             }
25         }
26         return productSum;
27     }
28 }

做法二：

 1 class SparseVector {
 2     List<int[]> list = new ArrayList<>();
 3     SparseVector(int[] nums) {
 4         for (int i = 0; i < nums.length; i++)
 5             if (nums[i] != 0)
 6                 list.add(new int[] {i, nums[i]});
 7     }
 8 
 9     public int dotProduct(SparseVector vec) {
10         List<int[]> passedIn = vec.list;
11         if (list.size() == 0 || passedIn.size() == 0) {
12             return 0;
13         }
14             
15         if (list.size() > passedIn.size()) {
16             return vec.dotProduct(this);
17         }
18             
19         int productSum = 0;
20         int idxCurrent = 0, idxPassedIn = 0;
21         while (idxCurrent < list.size() && idxPassedIn < passedIn.size()) {
22             int[] current = list.get(idxCurrent);
23             int[] passedInArray = passedIn.get(idxPassedIn);
24             
25             if (current[0] == passedInArray[0]) {
26                 productSum += current[1] * passedInArray[1];
27                 idxCurrent++;
28                 idxPassedIn++;
29             } else if (current[0] < passedInArray[0]) {
30                 idxCurrent++;
31             } else {
32                 idxPassedIn++;
33             }
34         }
35         return productSum;
36     }
37 }

做法三：

 1 class SparseVector {
 2     List<int[]> list = new ArrayList<>();
 3     SparseVector(int[] nums) {
 4         for (int i = 0; i < nums.length; i++)
 5             if (nums[i] != 0)
 6                 list.add(new int[] {i, nums[i]});
 7     }
 8 
 9     public int dotProduct(SparseVector vec) {
10         List<int[]> passedIn = vec.list;
11         if (list.size() == 0 || passedIn.size() == 0) {
12             return 0;
13         }
14             
15         if (list.size() > passedIn.size()) {
16             return vec.dotProduct(this);
17         }
18             
19         int productSum = 0;
20         for(int idxCurrent = 0; idxCurrent < list.size(); idxCurrent++) {
21             int[] current = list.get(idxCurrent);
22             int idxInPassedIn = binarySearch(passedIn, current[0]);
23             if (idxInPassedIn != -1) {
24                 productSum += current[1] * passedIn.get(idxInPassedIn)[1];
25             }
26         }
27         return productSum;
28     }
29     
30     private int binarySearch(List<int[]> list, int target) {
31         if (list == null || list.isEmpty()) {
32             return -1;
33         }
34         
35         int begin = 0;
36         int end =  list.size() - 1;
37         while (begin <= end) {
38             int mid = (end - begin) / 2 + begin;
39             if (list.get(mid)[0] == target) {
40                 return mid;
41             } else if (list.get(mid)[0] > target) {
42                 end = mid - 1;
43             } else {
44                 begin = mid + 1;
45             }
46         }
47         return -1;
48     }
49 }

 

posted @ 2021-03-12 05:53  北叶青藤  阅读(494)  评论(0)    收藏  举报