本问题对应的 leetcode 原文链接：剑指 Offer 31. 栈的压入、弹出序列
问题描述
输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。例如，序列 {1,2,3,4,5} 是某栈的压栈序列，序列 {4,5,3,2,1} 是该压栈序列对应的一个弹出序列，但 {4,3,5,1,2} 就不可能是该压栈序列的弹出序列。
示例1 ：
输入：pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
输出：true
解释：我们可以按以下顺序执行：
push(1), push(2), push(3), push(4), pop() -> 4,
push(5), pop() -> 5, pop() -> 3, pop() -> 2, pop() -> 1
示例2 ：
输入：pushed = [1,2,3,4,5], popped = [4,3,5,1,2]
输出：false
解释：1 不能在 2 之前弹出。
限制：
0 <= pushed.length == popped.length <= 10000 <= pushed[i], popped[i] < 1000pushed是popped的排列。
解题思路
视频讲解直达： 本题视频讲解
代码实现
时间复杂度：O(n)
额外空间复杂度：O(n)
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
if(pushed == null || pushed.length <= 0){
return true;
}
int k = 0;
Stack<Integer> stack = new Stack();
for(int i = 0; i < pushed.length; i++){
stack.push(pushed[i]);
while(!stack.isEmpty() && stack.peek() == popped[k]){
stack.pop();
k++;
}
}
return stack.isEmpty();
}
}
Python
class Solution(object):
def validateStackSequences(self, pushed, popped):
"""
:type pushed: List[int]
:type popped: List[int]
:rtype: bool
"""
if not pushed:
return True
k = 0
stack = []
for i in range(len(pushed)):
stack.append(pushed[i])
while stack and stack[-1] == popped[k]:
stack.pop()
k += 1
return not stack
C++
class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
if (pushed.empty()) {
return true;
}
int k = 0;
stack<int> s;
for (int i = 0; i < pushed.size(); ++i) {
s.push(pushed[i]);
while (!s.empty() && s.top() == popped[k]) {
s.pop();
++k;
}
}
return s.empty();
}
};
Go
func validateStackSequences(pushed []int, popped []int) bool {
if len(pushed) == 0 {
return true
}
k := 0
stack := []int{}
for _, v := range pushed {
stack = append(stack, v)
for len(stack) > 0 && stack[len(stack)-1] == popped[k] {
stack = stack[:len(stack)-1]
k++
}
}
return len(stack) == 0
}
JS
/**
* @param {number[]} pushed
* @param {number[]} popped
* @return {boolean}
*/
var validateStackSequences = function(pushed, popped) {
if (pushed == null || pushed.length <= 0) {
return true;
}
var k = 0;
var stack = [];
for (var i = 0; i < pushed.length; i++) {
stack.push(pushed[i]);
while (stack.length > 0 && stack[stack.length - 1] == popped[k]) {
stack.pop();
k++;
}
}
return stack.length == 0;
};
shuaidi