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 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 <= stones.length <= 30
- 1 <= stones[i] <= 1000
方法1: multiset
class Solution { public: int lastStoneWeight(vector<int>& stones) { multiset<int> st(stones.begin(), stones.end()); while (st.size() >= 2) { int mx = *st.rbegin(); st.erase(st.find(mx)); int nx = *st.rbegin(); st.erase(st.find(nx)); st.insert(mx - nx); } return st.empty() ? 0 : *st.begin(); } };