Implement a SnapshotArray that supports the following interface:
SnapshotArray(int length)
initializes an array-like data structure with the given length. Initially, each element equals 0.void set(index, val)
sets the element at the givenindex
to be equal toval
.int snap()
takes a snapshot of the array and returns thesnap_id
: the total number of times we calledsnap()
minus1
.int get(index, snap_id)
returns the value at the givenindex
, at the time we took the snapshot with the givensnap_id
Example 1:
Input: ["SnapshotArray","set","snap","set","get"] [[3],[0,5],[],[0,6],[0,0]] Output: [null,null,0,null,5] Explanation: SnapshotArray snapshotArr = new SnapshotArray(3); // set the length to be 3 snapshotArr.set(0,5); // Set array[0] = 5 snapshotArr.snap(); // Take a snapshot, return snap_id = 0 snapshotArr.set(0,6); snapshotArr.get(0,0); // Get the value of array[0] with snap_id = 0, return 5
Constraints:
1 <= length <= 50000
- At most
50000
calls will be made toset
,snap
, andget
. 0 <= index < length
0 <= snap_id <
(the total number of times we callsnap()
)0 <= val <= 10^9
Solution: map + upper_bound
Use a vector to store maps, one map per element.
The map stores {snap_id -> val}, use upper_bound to find the first version > snap_id and use previous version’s value.
Time complexity:
Set: O(log|snap_id|)
Get: O(log|snap_id|)
Snap: O(1)
Space complexity: O(length + set_calls)
C++
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// Author: Huahua class SnapshotArray { public: SnapshotArray(int length): id_(0), vals_(length) {} void set(int index, int val) { vals_[index][id_] = val; } int snap() { return id_++; } int get(int index, int snap_id) const { auto it = vals_[index].upper_bound(snap_id); if (it == begin(vals_[index])) return 0; return prev(it)->second; } private: int id_; vector<map<int, int>> vals_; }; |
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