Last Stone Weight

Similar Problems:

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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

- Solution:

// https://code.dennyzhang.com/last-stone-weight // Basic Ideas: heap // Complexity: Time O(n*log(n)), Space O(n) import ( "container/heap" "fmt" ) // An IntHeap is a max-heap of ints. type IntHeap []int func (h IntHeap) Len() int { return len(h) } func (h IntHeap) Less(i, j int) bool { return h[i] > h[j] } func (h IntHeap) Swap(i, j int) { h[i], h[j] = h[j], h[i] } func (h *IntHeap) Push(x interface{}) { // Push and Pop use pointer receivers because they modify the slice's length, // not just its contents. *h = append(*h, x.(int)) } func (h *IntHeap) Pop() interface{} { old := *h n := len(old) x := old[n-1] *h = old[0 : n-1] return x } func lastStoneWeight(stones []int) int { h := &IntHeap{} heap.Init(h) for _, val := range stones { heap.Push(h, val) } for h.Len() > 1 { v1, _ := heap.Pop(h).(int) v2, _ := heap.Pop(h).(int) heap.Push(h, v1-v2) } res, _ := heap.Pop(h).(int) return res }