Sword finger offer 31 Stack push in and pop-up sequence

subject

Enter two sequences of integers , The first sequence represents the push order of the stack , Determine whether the second sequence is the pop-up order for the stack . Suppose that all the Numbers pushed are not equal . for example , Sequence {1,2,3,4,5} Is the stack pressing sequence of a stack , Sequence {4,5,3,2,1} Is a popup sequence corresponding to the stack sequence , but {4,3,5,1,2} It cannot be a popup sequence of the push sequence .、

Example 1：

 Input ：pushed = [1,2,3,4,5], popped = [4,5,3,2,1]
 Output ：true
 explain ： We can do it in the following order ：
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
 explain ：1  Can't be in  2  Pop up before .

Tips ：

0 <= pushed.length == popped.length <= 1000
0 <= pushed[i], popped[i] < 1000
pushed  yes  popped  Permutation .

 

public static boolean isStackArray(int[] nums1, int[] nums2) {
        if (nums1.length != nums2.length) {
            return false;
        }

        int index1 = nums1.length - 1;
        LinkedList<Integer> stack = new LinkedList<>();
        for (int index2 = 0; index2 < nums2.length; index2++) {
            stack.push(nums2[index2]);

            while (!stack.isEmpty() && nums1[index1] == stack.peek()) {
                index1--;
                stack.pop();
            }
        }

        return stack.isEmpty();
    }