## 1046. Last Stone Weight

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:1Explanation: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.

**Note:**

`1 <= stones.length <= 30`

`1 <= stones[i] <= 1000`

## Rust Solution

```
struct Solution;
use std::collections::BinaryHeap;
impl Solution {
fn last_stone_weight(stones: Vec<i32>) -> i32 {
let mut pq: BinaryHeap<i32> = BinaryHeap::from(stones);
while let Some(a) = pq.pop() {
if let Some(b) = pq.pop() {
if a - b != 0 {
pq.push(a - b);
}
} else {
return a;
}
}
0
}
}
#[test]
fn test() {
let stones = vec![2, 7, 4, 1, 8, 1];
assert_eq!(Solution::last_stone_weight(stones), 1);
}
```

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