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.

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        int n = pushed.length;
        int j = 0;
        Stack<Integer> stack = new Stack();
        
        for(int num : pushed) {
            stack.push(num);
            while(!stack.isEmpty() && stack.peek() == popped[j]) {
                stack.pop();
                j++;
                
            }
        }
        return j == n;
    }
}

greedy, 当栈顶等于popped时pop出，同时j++，最后如果所有数字都match上了就说明是对的。