题目如下:
Given two sequences
pushed
andpopped
with distinct values, returntrue
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 = [4,5,3,2,1] 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 = [1,2,3,4,5], popped = [4,3,5,1,2] 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 ofpopped
.pushed
andpopped
have distinct values.
解题思路:栈具有后进先出的特性,从这个特性中可以得出这么一个规律:对于出栈序列中的任意一个元素来说,在这个元素之前入栈的其他元素并且在这个元素之后出栈的这些元素必定满足后进先出。以 pushed = [8,2,1,4,7,9,0,3,5,6],popped = [1,2,7,3,6,4,0,9,5,8] 为例,对于元素3来说,在其之前入栈的元素为[8,2,1,4,7,9,0],在其后出栈的元素为[6,4,0,9,5,8] ,同时满足先于3入栈后于3出栈的元素为[8,4,9,0],这些元素出栈的顺序必定要满足[0,9,4,8]的顺序。所以只要遍历[8,4,9,0]序列,判断这些元素的在popped数组中的索引是否是递减即可。
代码如下:
class Solution(object): def validateStackSequences(self, pushed, popped): """ :type pushed: List[int] :type popped: List[int] :rtype: bool """ dic_push = {} dic_pop = {} for i,v in enumerate(pushed): dic_push[v] = i for i, v in enumerate(popped): dic_pop[v] = i for i,v in enumerate(popped): push_inx = dic_push[v] lastInx = len(popped) for j in range(push_inx): if dic_pop[pushed[j]] < i: continue elif dic_pop[pushed[j]] > lastInx : return False lastInx = dic_pop[pushed[j]] return True