You have some number of sticks with positive integer lengths. These lengths are given as an array sticks, where sticks[i] is the length of the ith stick.

You can connect any two sticks of lengths x and y into one stick by paying a cost of x + y. You must connect all the sticks until there is only one stick remaining.

Return *the minimum cost of connecting all the given sticks into one stick in this way*.

Input:sticks = [2,4,3]Output:14Explanation:You start with sticks = [2,4,3]. 1. Combine sticks 2 and 3 for a cost of 2 + 3 = 5. Now you have sticks = [5,4]. 2. Combine sticks 5 and 4 for a cost of 5 + 4 = 9. Now you have sticks = [9]. There is only one stick left, so you are done. The total cost is 5 + 9 = 14.

Input:sticks = [1,8,3,5]Output:30Explanation:You start with sticks = [1,8,3,5]. 1. Combine sticks 1 and 3 for a cost of 1 + 3 = 4. Now you have sticks = [4,8,5]. 2. Combine sticks 4 and 5 for a cost of 4 + 5 = 9. Now you have sticks = [9,8]. 3. Combine sticks 9 and 8 for a cost of 9 + 8 = 17. Now you have sticks = [17]. There is only one stick left, so you are done. The total cost is 4 + 9 + 17 = 30.

Input:sticks = [5]Output:0Explanation:There is only one stick, so you don't need to do anything. The total cost is 0.

- 1 <= sticks.length <= 104
- 1 <= sticks[i] <= 104

Solution:

class Solution { public int connectSticks(int[] sticks) { PriorityQueue<Integer> pq = new PriorityQueue(); for (int s : sticks) pq.offer(s); int cost = 0; while (pq.size() > 1) { int first = pq.poll(); int second = pq.poll(); cost += first + second; pq.offer(first + second); } return cost; } }