Class Solution
Easy
You are given an array of integers stones where stones[i] is the weight of the ith stone.
We are playing a game with the stones. On each turn, we choose the heaviest two stones and smash them together. Suppose the heaviest two stones have weights x and y with x <= y. The result of this smash is:
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x == y, both stones are destroyed, and - If
x != y, the stone of weightxis destroyed, and the stone of weightyhas new weighty - x.
At the end of the game, there is at most one stone left.
Return the smallest possible weight of the left stone. If there are no stones left, return 0.
Example 1:
Input: stones = [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 the last stone.
Example 2:
Input: stones = [1]
Output: 1
Constraints:
1 <= stones.length <= 301 <= stones[i] <= 1000
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Constructor Summary
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Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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lastStoneWeight
public int lastStoneWeight(int[] stones)
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