# 1049. Last Stone Weight II

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 any two 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 destroyed, and
• If `x != y`, the stone of weight `x` is destroyed, and the stone of weight `y` has new weight `y - 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 can combine 2 and 4 to get 2, so the array converts to [2,7,1,8,1] then,
we can combine 7 and 8 to get 1, so the array converts to [2,1,1,1] then,
we can combine 2 and 1 to get 1, so the array converts to [1,1,1] then,
we can combine 1 and 1 to get 0, so the array converts to [1], then that's the optimal value.
``````

Example 2:

``````Input: stones = [31,26,33,21,40]
Output: 5
``````

Example 3:

``````Input: stones = [1,2]
Output: 1
``````

Constraints:

• `1 <= stones.length <= 30`
• `1 <= stones[i] <= 100`

``````class Solution {
public:
int lastStoneWeightII(vector<int>& stones) {
vector<bool> dp(1501);
dp[0] = true;
int sum = 0;
for (int stone : stones) {
sum += stone;
for (int i = min(1500, sum); i >= stone; --i) {
dp[i] = dp[i] || dp[i - stone];
}
}
for (int i = sum / 2; i >= 0; --i) {
if (dp[i]) return sum - 2 * i;
}
return 0;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1049

Last Stone Weight

Coin Change

https://leetcode.com/problems/last-stone-weight-ii/

https://leetcode.com/problems/last-stone-weight-ii/discuss/294888/JavaC%2B%2BPython-Easy-Knapsacks-DP

https://leetcode.com/problems/last-stone-weight-ii/discuss/295167/Java-beat-100-with-nice-explanation

LeetCode All in One 题目讲解汇总(持续更新中…)

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