Description
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 <= 5 * 104
0 <= index < length
0 <= val <= 109
0 <= snap_id <
(the total number of times we callsnap()
)- At most
5 * 104
calls will be made toset
,snap
, andget
.
Code
Time Complexity: , Space Complexity:
建立 base case:
for(int i = 0; i < length; i++) {
map<int, int> mp;
mp[0] = 0;
A[i] = mp;
}
如此一來在 call
return prev(A[index].upper_bound(snap_id))->second;
時就不用擔心 prev
之後會沒有值可以取(map 有可能是空的)。
class SnapshotArray {
public:
vector<map<int, int>> A;
int snap_id;
SnapshotArray(int length) {
A.resize(length);
for(int i = 0; i < length; i++) {
map<int, int> mp;
mp[0] = 0;
A[i] = mp;
}
snap_id = 0;
}
void set(int index, int val) {
A[index][snap_id] = val;
}
int snap() {
return snap_id++;
}
int get(int index, int snap_id) {
return prev(A[index].upper_bound(snap_id))->second;
}
};
/**
* Your SnapshotArray object will be instantiated and called as such:
* SnapshotArray* obj = new SnapshotArray(length);
* obj->set(index,val);
* int param_2 = obj->snap();
* int param_3 = obj->get(index,snap_id);
*/