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: 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.

 

Note:

  1. 1 <= stones.length <= 30
  2. 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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