0946. Validate Stack Sequences (M)

题目

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.

Constraints:

• 0 <= pushed.length == popped.length <= 1000
• 0 <= pushed[i], popped[i] < 1000
• pushed is a permutation of popped.
• pushed and popped have distinct values.

题意

给定一个栈的入栈顺序表和出栈顺序表，判断能否实现对应的出入栈操作(符合先进后出)。

思路

贪心+模拟。维护一个栈，当栈顶元素和出栈表的下一个元素相同时，立即出栈；否则压入入栈表的下一个元素。反证法证明正确性：栈中元素都不相同，记当前出栈表的第一个元素为x，如果栈顶元素也为x，且选择不出栈，那么x必然不会成为第一个出栈的元素，矛盾。

代码实现

Java

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Deque<Integer> stack = new ArrayDeque<>();
        int i = 0, j = 0;
        while (j < popped.length) {
            if (i == pushed.length && stack.peek() != popped[j]) return false;
            if (stack.isEmpty() || stack.peek() != popped[j]) {
                stack.push(pushed[i++]);
            } else {
                stack.pop();
                j++;
            }
        }
        return true;
    }
}