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Last Stone Weight.cpp
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Last Stone Weight.cpp
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Task: 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 <= stones.length <= 30
1 <= stones[i] <= 1000
Solution:
class Solution {
public:
int lastStoneWeight(vector<int>& stones) {
while(stones.size()!=1)
{
sort(stones.begin(),stones.end());
int x=stones[stones.size()-2];
int y=stones[stones.size()-1];
if(x==y)
{
stones.erase(stones.end()-2);
stones.erase(stones.end()-1);
if(stones.size()==0)
{
return 0;
break;
}
}
else
{
stones.erase(stones.end()-2);
stones.erase(stones.end()-1);
y=y-x;
stones.push_back(y);
}
}
return stones[0];
}
};