1046. Last Stone Weight
We have a collection of rocks, each rock has a positive integer weight.
Each turn, we choose the two heaviest rocks 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 weightx
is totally destroyed, and the stone of weighty
has new weighty-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 <= stones.length <= 30
1 <= stones[i] <= 1000
int lastStoneWeight(vector<int>& stones) { // time: O(nlogn); space: O(n)
priority_queue<int> pq(stones.begin(), stones.end());
int x, y;
while (pq.size() > 1) {
y = pq.top(); pq.pop();
x = pq.top(); pq.pop();
if (y > x) pq.push(y - x);
}
return pq.empty() ? 0 : pq.top();
}
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