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Merge pull request soapyigu#235 from iHackSubhodip/master
Optimize solution to Decode String & LRU Cache
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/** | ||
* Question Link: https://leetcode.com/problems/lru-cache/ | ||
* Primary idea: Use Doubly linked list and hash table to build the LRU cache. | ||
* Time Complexity: O(1), Space Complexity: O(n) | ||
* | ||
*/ | ||
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class DoublyLinkedList{ | ||
var key: Int | ||
var value: Int | ||
var previous: DoublyLinkedList? | ||
var next: DoublyLinkedList? | ||
var hashValue: Int | ||
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init(_ key: Int, _ value: Int) { | ||
self.key = key | ||
self.value = value | ||
self.hashValue = key | ||
} | ||
} | ||
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class LRUCache{ | ||
var maxCapacity: Int | ||
var head: DoublyLinkedList | ||
var tail: DoublyLinkedList | ||
var cache: [Int: DoublyLinkedList] | ||
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init(_ maxCapacity: Int) { | ||
self.maxCapacity = maxCapacity | ||
self.cache = [Int: DoublyLinkedList]() | ||
self.head = DoublyLinkedList(Int.min, Int.min) | ||
self.tail = DoublyLinkedList(Int.max, Int.max) | ||
self.head.next = self.tail | ||
self.tail.previous = self.head | ||
} | ||
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func add(_ node: DoublyLinkedList){ | ||
let next = head.next | ||
head.next = node | ||
node.previous = head | ||
node.next = next | ||
next?.previous = node | ||
} | ||
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func remove(_ node: DoublyLinkedList){ | ||
let previous = node.previous | ||
let next = node.next | ||
previous?.next = next | ||
next?.previous = previous | ||
} | ||
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func get(_ key: Int) -> Int{ | ||
if let node = cache[key]{ | ||
remove(node) | ||
add(node) | ||
return node.value | ||
} | ||
return -1 | ||
} | ||
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func put(_ key: Int, _ value: Int){ | ||
if let node = cache[key]{ | ||
remove(node) | ||
cache.removeValue(forKey: key) | ||
}else if cache.keys.count >= maxCapacity{ | ||
if let leastNode = tail.previous{ | ||
remove(leastNode) | ||
cache.removeValue(forKey: leastNode.key) | ||
} | ||
} | ||
let newNode = DoublyLinkedList(key, value) | ||
cache[key] = newNode | ||
add(newNode) | ||
} | ||
} |
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Original file line number | Diff line number | Diff line change |
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@@ -0,0 +1,52 @@ | ||
/** | ||
* Question Link: https://leetcode.com/problems/decode-string/ | ||
* Primary idea: Primary idea is to maintain two stacks[i.e. countStack and characterStack]. | ||
* Traverse the given string and process the elements into the respective stacks, and accordingly update the result. | ||
* Time Complexity: O(n), Space Complexity: O(n) | ||
* | ||
*/ | ||
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class Solution { | ||
func decodeString(_ s: String) -> String { | ||
var result = "" | ||
var countStack = [Int]() | ||
var characterStack = [String]() | ||
let allowedDigits = Set(["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]) | ||
var arrayString = Array(s), i = 0 | ||
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while i < arrayString.count{ | ||
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if allowedDigits.contains(String(arrayString[i])){ | ||
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var count = 0 | ||
while allowedDigits.contains(String(arrayString[i])){ | ||
count = 10 * count + Int(String(arrayString[i]))! | ||
i += 1 | ||
} | ||
countStack.append(count) | ||
}else if arrayString[i] == "["{ | ||
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characterStack.append(result) | ||
result = "" | ||
i += 1 | ||
}else if arrayString[i] == "]"{ | ||
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if var temp = characterStack.popLast(), let repeatTime = countStack.popLast(){ | ||
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for _ in 0..<repeatTime{ | ||
temp.append(result) | ||
} | ||
result = temp | ||
} | ||
i += 1 | ||
}else{ | ||
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result.append(arrayString[i]) | ||
i += 1 | ||
} | ||
} | ||
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return result | ||
} | ||
} |