LeetCode 1146. Snapshot Array

标签：object   xpl   which   value   structure   nbsp   public   space   ems   

原题链接在这里：https://leetcode.com/problems/snapshot-array/

题目：

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 given index to be equal to val.
• int snap() takes a snapshot of the array and returns the snap_id: the total number of times we called snap() minus 1.
• int get(index, snap_id) returns the value at the given index, at the time we took the snapshot with the given snap_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 to set, snap, and get.
• 0 <= index < length
• 0 <= snap_id < (the total number of times we call snap())
• 0 <= val <= 10^9

题解：

Instead of make a copy of each snapshot, which takes a lot of memory space, we could record the state of cell when calling set method.

Have a TreeMap array, each TreeMap maintains the states of a cell.

When calling set, mark current snapshot id with the new value of this cell.

When calling get, try to get the floor entry with given snapshot id.

Time Complexity: SnapshotArray, O(length). set, O(logn). snap, O(1). get, O(logn). n is the number of total entries in arr, the number of previoius set call.

Space: O(n). 

AC Java:

 1 class SnapshotArray {
 2     TreeMap<Integer, Integer> [] arr;
 3     int snapId;
 4     
 5     public SnapshotArray(int length) {
 6         arr = new TreeMap[length];
 7         for(int i = 0; i<length; i++){
 8             arr[i] = new TreeMap<Integer, Integer>();
 9             arr[i].put(0, 0);
10         }
11         
12         snapId = 0;
13     }
14     
15     public void set(int index, int val) {
16         arr[index].put(snapId, val);
17     }
18     
19     public int snap() {
20         return snapId++;
21     }
22     
23     public int get(int index, int snap_id) {
24         return arr[index].floorEntry(snap_id).getValue();
25     }
26 }
27 
28 /**
29  * Your SnapshotArray object will be instantiated and called as such:
30  * SnapshotArray obj = new SnapshotArray(length);
31  * obj.set(index,val);
32  * int param_2 = obj.snap();
33  * int param_3 = obj.get(index,snap_id);
34  */

LeetCode 1146. Snapshot Array

标签：object   xpl   which   value   structure   nbsp   public   space   ems   

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/12041830.html