给定两个稀疏向量,计算它们的点积(数量积)。
实现类 SparseVector
:
SparseVector(nums)
以向量nums
初始化对象。dotProduct(vec)
计算此向量与vec
的点积。
稀疏向量 是指绝大多数分量为 0 的向量。你需要 高效 地存储这个向量,并计算两个稀疏向量的点积。
进阶:当其中只有一个向量是稀疏向量时,你该如何解决此问题?
示例 1:
输入:nums1 = [1,0,0,2,3], nums2 = [0,3,0,4,0] 输出:8 解释:v1 = SparseVector(nums1) , v2 = SparseVector(nums2) v1.dotProduct(v2) = 1*0 + 0*3 + 0*0 + 2*4 + 3*0 = 8
示例 2:
输入:nums1 = [0,1,0,0,0], nums2 = [0,0,0,0,2] 输出:0 解释:v1 = SparseVector(nums1) , v2 = SparseVector(nums2) v1.dotProduct(v2) = 0*0 + 1*0 + 0*0 + 0*0 + 0*2 = 0
示例 3:
输入:nums1 = [0,1,0,0,2,0,0], nums2 = [1,0,0,0,3,0,4] 输出:6
提示:
n == nums1.length == nums2.length
1 <= n <= 10^5
0 <= nums1[i], nums2[i] <= 100
哈希表实现。
用哈希表存储非 0 点的下标与值。求点积时,遍历长度较小的哈希表。
class SparseVector:
def __init__(self, nums: List[int]):
self.v = {}
for i, num in enumerate(nums):
if num != 0:
self.v[i] = num
# Return the dotProduct of two sparse vectors
def dotProduct(self, vec: 'SparseVector') -> int:
res = 0
if len(self.v) > len(vec.v):
self.v, vec.v = vec.v, self.v
for i, num in self.v.items():
if i not in vec.v:
continue
res += num * vec.v[i]
return res
# Your SparseVector object will be instantiated and called as such:
# v1 = SparseVector(nums1)
# v2 = SparseVector(nums2)
# ans = v1.dotProduct(v2)
class SparseVector {
private Map<Integer, Integer> v;
SparseVector(int[] nums) {
v = new HashMap<>();
for (int i = 0; i < nums.length; ++i) {
if (nums[i] != 0) {
v.put(i, nums[i]);
}
}
}
// Return the dotProduct of two sparse vectors
public int dotProduct(SparseVector vec) {
int res = 0;
if (v.size() > vec.v.size()) {
Map<Integer, Integer> t = v;
v = vec.v;
vec.v = t;
}
for (Map.Entry<Integer, Integer> entry : v.entrySet()) {
int i = entry.getKey(), num = entry.getValue();
res += num * vec.v.getOrDefault(i, 0);
}
return res;
}
}
// Your SparseVector object will be instantiated and called as such:
// SparseVector v1 = new SparseVector(nums1);
// SparseVector v2 = new SparseVector(nums2);
// int ans = v1.dotProduct(v2);