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
public class Solution {
int[] heap;
int index = 1;
public int LastStoneWeight(int[] stones)
{
if(stones.Length==0){
return 0;
}
if(stones.Length==1){
return stones[0];
}
heap = new int[stones.Length + 1];
for (int i = 0; i < stones.Length; i++)
{
Insert(stones[i]);
}
int diff = 0;
while (index-1 > 1)
{
int max1 = ExtractMax();
int max2 = ExtractMax();
diff = max1 - max2;
Insert(diff);
}
return diff;
}
private int ExtractMax()
{
int max = heap[1];
index--;
heap[1] = heap[index];
HeapifyDown(1);
return max;
}
private void HeapifyDown(int idx)
{
int left = 2 * idx;
int right = 2 * idx + 1;
if (left > index)
{
return;
}
if (right > index)
{
if (heap[left] > heap[idx])
{
int temp = heap[left];
heap[left] = heap[idx];
heap[idx] = temp;
HeapifyDown(left);
}
return;
}
if (heap[left] > heap[right] && heap[left] > heap[idx])
{
int temp = heap[left];
heap[left] = heap[idx];
heap[idx] = temp;
HeapifyDown(left);
}
else if (heap[right] > heap[idx])
{
int temp = heap[right];
heap[right] = heap[idx];
heap[idx] = temp;
HeapifyDown(right);
}
}
private void Insert(int val)
{
heap[index] = val;
HeapifyUp(index);
index++;
}
private void HeapifyUp(int idx)
{
int parent = idx / 2;
if (parent < 1)
{
return;
}
if (heap[parent] < heap[idx])
{
int temp = heap[parent];
heap[parent] = heap[idx];
heap[idx] = temp;
HeapifyUp(parent);
}
}
}
Time Complexity: O(nlogn)
Space Complexity: O(n)