Formatted question description: https://leetcode.ca/all/946.html
946. Validate Stack Sequences (Medium)
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.
Related Topics:
Stack
Solution 1.
// OJ: https://leetcode.com/problems/validate-stack-sequences/
// Time: O(N)
// Space: O(N)
class Solution {
public:
bool validateStackSequences(vector<int>& pushed, vector<int>& popped) {
stack<int> s;
int p = 0;
for (int i : pushed) {
s.push(i);
while (s.size() && s.top() == popped[p]) {
s.pop();
++p;
}
}
return s.empty();
}
};
Java
-
class Solution { public boolean validateStackSequences(int[] pushed, int[] popped) { Stack<Integer> stack = new Stack<Integer>(); int length = pushed.length; int pushIndex = 0, popIndex = 0; while (pushIndex < length && popIndex < length) { stack.push(pushed[pushIndex]); pushIndex++; while (!stack.isEmpty() && stack.peek() == popped[popIndex]) { stack.pop(); popIndex++; } } return stack.isEmpty(); } }
-
// OJ: https://leetcode.com/problems/validate-stack-sequences/ // Time: O(N) // Space: O(N) class Solution { public: bool validateStackSequences(vector<int>& pushed, vector<int>& popped) { stack<int> s; int p = 0; for (int i : pushed) { s.push(i); while (s.size() && s.top() == popped[p]) { s.pop(); ++p; } } return s.empty(); } };
-
class Solution(object): def validateStackSequences(self, pushed, popped): """ :type pushed: List[int] :type popped: List[int] :rtype: bool """ stack = [] N = len(pushed) pi = 0 for i in range(N): if stack and popped[i] == stack[-1]: stack.pop() else: while pi < N and pushed[pi] != popped[i]: stack.append(pushed[pi]) pi += 1 pi += 1 return not stack