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 ofpopped
.pushed
andpopped
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;
}
}