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.
Example 1:
Input: sticks = [2,4,3] Output: 14 Explanation: 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.
Example 2:
Input: sticks = [1,8,3,5] Output: 30 Explanation: 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.
Example 3:
Input: sticks = [5] Output: 0 Explanation: There is only one stick, so you don't need to do anything. The total cost is 0.
Constraints:
1 <= sticks.length <= 1041 <= sticks[i] <= 104
题目给定一个长度为n的数组sticks表示有n根棍子，sticks[i]表示第i根棍子的长度。每次可以任意挑选两根棍子连在一起，假定它们的长度分别为x和y，那么花费就是x+y。现在要求以最少的花费把所有棍子连成一根棍子，问最少花费应该是多少。
根据题意，花费是由棍子长度决定的，棍子越长花费越多。另外两根棍子连在一起后，还需要跟后面的棍子连在一起，也就是说先被挑中连接的棍子，将会在后面再次跟其它棍子相连接，它们的花费也将会被累加。本着贪心法则，我们当然希望长的棍子花费尽量少地被累加，因此我们应该每次先挑最短的两根相连，尽可能地把长的棍子留到后面。
总是要从一组数中挑选最小值，最容易想到的就是优先级队列（最小堆）。把所有棍子的长度值放入一个最小堆中，那么每次从最小堆弹出的两个就是最短的两根棍子，连在一起后把总花费计入结果，然后再把连在一起的新棍子的长度值放回到最小堆里。重复这样操作，直到最小堆里只剩下一个值，即所有棍子都被连在了一起形成一根长棍子。
class Solution:
def connectSticks(self, sticks :List[int]) -> int:
res = 0
heapq.heapify(sticks)
while len(sticks) > 1 :
s = heapq.heappop(sticks)
s += heapq.heappop(sticks)
heapq.heappush(sticks, s)
res += s
return res
