modify solution to use deque data structure instead of stack to better align with book solution

java/Stacks/MaximumsOfSlidingWindow.java

@@ -1,28 +1,29 @@
-import java.util.Stack;
+import java.util.ArrayDeque;
+import java.util.Deque;
 
 public class MaximumsOfSlidingWindow {
     public int[] maximumsOfSlidingWindow(int[] nums, int k) {
         int[] res = new int[nums.length - k + 1];
-        Stack<int[]> dq = new Stack<>();
+        Deque<int[]> dq = new ArrayDeque<>();
         int left, right;
         left = right = 0;
         while (right < nums.length) {
             // 1) Ensure the values of the deque maintain a monotonic decreasing order
             // by removing candidates <= the current candidate.
-            while (!dq.isEmpty() && dq.peek()[0] <= nums[right]) {
-                dq.pop();
+            while (!dq.isEmpty() && dq.peekLast()[0] <= nums[right]) {
+                dq.pollLast();
             }
             // 2) Add the current candidate.
-            dq.push(new int[]{nums[right], right});
+            dq.offerLast(new int[]{nums[right], right});
             // If the window is of length 'k', record the maximum of the window.
             if (right - left + 1 == k) {
                 // 3) Remove values whose indexes occur outside the window.
-                if (!dq.isEmpty() && dq.firstElement()[1] < left) {
-                    dq.remove(0);
+                if (!dq.isEmpty() && dq.peekFirst()[1] < left) {
+                    dq.pollFirst();
                 }
                 // The maximum value of this window is the leftmost value in the 
                 // deque.
-                res[left] = dq.firstElement()[0];
+                res[left] = dq.peekFirst()[0];
                 // Slide the window by advancing both 'left' and 'right'. The right  
                 // pointer always gets advanced so we just need to advance 'left'.
                 left++;