We have a collection of stones, each stone has a positive integer weight.

Each turn, we choose the two **heaviest** stones and smash them together. Suppose the stones have weights `x`

and `y`

with `x <= y`

. The result of this smash is:

- If
`x == y`

, both stones are totally destroyed; - If
`x != y`

, the stone of weight`x`

is totally destroyed, and the stone of weight`y`

has new weight`y-x`

.

At the end, there is at most 1 stone left. Return the weight of this stone (or 0 if there are no stones left.)

```
Example 1:
Input: [2,7,4,1,8,1]
Output: 1
Explanation:
We combine 7 and 8 to get 1 so the array converts to [2,4,1,1,1] then,
we combine 2 and 4 to get 2 so the array converts to [2,1,1,1] then,
we combine 2 and 1 to get 1 so the array converts to [1,1,1] then,
we combine 1 and 1 to get 0 so the array converts to [1] then that's the value of last stone.
```

Solution:

```
import "container/heap"
type IntHeap []int
func (h IntHeap) Len() int {return len(h)}
func (h IntHeap) Less(x, y int) bool {return h[x] > h[y]}
func (h IntHeap) Swap(x, y int) {h[x], h[y] = h[y], h[x]}
func (h *IntHeap) Push(x interface{}) {
*h = append(*h, x.(int))
}
func (h *IntHeap) Pop() interface{} {
if len(*h) == 0 {
return 0
}
old := *h
x := old[len(old) - 1]
*h = old[0: len(old) - 1]
return x
}
func lastStoneWeight(stones []int) int {
h := &IntHeap{}
heap.Init(h) //important
for i := 0; i < len(stones); i++ {
heap.Push(h, stones[i])
}
for len(*h) >= 2 {
a := (heap.Pop(h)).(int)
b := (heap.Pop(h)).(int)
if b > a{
b, a = a, b
}
heap.Push(h, a - b)
}
return (heap.Pop(h)).(int)
}
```