Given two sequences pushed
and popped
with distinct values, return true
if and only if this Could have been the result of a sequence of push and pop operations on an initially empty stack.
Example 1:
input: pushed = [1,2,3,4,5],popped = [4,5,1] Output: true Explanation: We might do the following sequence: push(1),push(2),push(3),push(4),pop() -> 4,push(5),pop() -> 5,pop() -> 3,pop() -> 2,pop() -> 1
Example 2:
input: pushed = [1,popped = [4,1,2] Output: false Explanation: 1 cannot be popped before 2.
Note:
0 <= pushed.length == popped.length <= 1000
0 <= pushed[i],popped[i] < 1000
pushed
is a permutation of popped
. pushed
and popped
have distinct values. 给定 pushed
和 popped
两个序列,只有当它们可能是在最初空栈上进行的推入 push 和d出 pop *** 作序列的结果时,返回 true
;否则,返回 false
。
示例 1:
输入:pushed = [1,popped = [4,1]输出:true解释:我们可以按以下顺序执行:push(1),pop() -> 1
示例 2:
输入:pushed = [1,2]输出:false解释:1 不能在 2 之前d出。
提示:
0 <= pushed.length == popped.length <= 1000
0 <= pushed[i],popped[i] < 1000
pushed
是 popped
的排列。 72ms 1 class Solution { 2 func valIDateStackSequences(_ pushed: [Int],_ popped: [Int]) -> Bool { 3 var s:Stack<Int> = Stack<Int>() 4 var j:Int = 0 5 for i in 0..<pushed.count 6 { 7 s.push(pushed[i]) 8 while (!s.isEmpty() && j < popped.count && s.getLast() == popped[j]) 9 {10 s.pop()11 j += 112 }13 }14 return s.isEmpty() && j == popped.count15 }16 }17 18 public struct Stack<T> {19 20 // 泛型数组:用于存储数据元素21 fileprivate var stack: [T] 22 23 // 返回堆栈中元素的个数24 public var count: Int {25 return stack.count26 }27 28 /// 构造函数:创建一个空的堆栈29 public init() {30 stack = [T]()31 }32 33 // 检查堆栈是否为空34 // - returns: 如果堆栈为空,则返回true,否则返回false35 public func isEmpty() -> Bool {36 return stack.isEmpty37 }38 39 // 入堆栈 *** 作:将元素添加到堆栈的末尾40 public mutating func push(_ element: T) {41 stack.append(element)42 }43 44 // 出堆栈 *** 作:删除并返回堆栈中的第一个元素45 public mutating func pop() -> T? {46 return stack.removeLast()47 }48 49 // 返回堆栈中的第一个元素(不删除)50 public func getLast() -> T? {51 return stack.last!52 }53 }总结
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