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.
思路:就是pushed直接push进stack,遇见poped,就pop,然后看是否能够最后是空stack。
class Solution {
public boolean validateStackSequences(int[] pushed, int[] popped) {
if(pushed == null || popped == null) {
return false;
}
Stack<Integer> stack = new Stack<Integer>();
int i = 0;
for(int x : pushed) {
stack.push(x);
while(!stack.isEmpty() && popped[i] == stack.peek()) {
stack.pop();
i++;
}
}
return stack.isEmpty();
}
}