[leetcode] 1146. Snapshot Array
原创
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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 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
分析
题目的意思是:实现一个镜像数组,我用list和字典暴力了一下,发现超时了。看了别人的实现后,发现在一开始可以不用申请数组,直接用字典搞定,然后在镜像的时候用数组存放镜像就行了,看来我正在处于并将长期处于被虐的阶段。
代码
class SnapshotArray:
def __init__(self, length: int):
self.arr={}
self.snap_id=0
self.snap_arr=[]
def set(self, index: int, val: int) -> None:
self.arr[index]=val
def snap(self) -> int:
self.snap_arr.append(dict(self.arr))
self.snap_id+=1
return self.snap_id-1
def get(self, index: int, snap_id: int) -> int:
arr=self.snap_arr[snap_id]
if(index in arr):
return arr[index]
else:
return 0
# Your SnapshotArray object will be instantiated and called as such:
# obj = SnapshotArray(length)
# obj.set(index,val)
# param_2 = obj.snap()
# param_3 = obj.get(index,snap_id)
参考文献
[LeetCode] Python 3 | Three Methods | Explanations