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 = [4,5,3,2,1] 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.
题意
给定两个序列push[]、popped[]。 用Stack验证用两个序列的元素,经过一番push和pop操作, Stack能为空
Solution: Stack
code
1 class Solution { 2 public boolean validateStackSequences(int[] pushed, int[] popped) { 3 Stack<Integer> stack = new Stack<>(); 4 int i = 0; 5 for (int p : pushed) { 6 stack.push(p); 7 while (!stack.isEmpty() && stack.peek() == popped[i]) { 8 stack.pop(); 9 ++i; 10 } 11 } 12 return stack.empty(); 13 } 14 }