LeetCode - Arrays - Last Stone Weight 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. class Solution { public int lastStoneWeight(int[] stones) { List<Integer> stone_list = Arrays.stream(stones).boxed().collect(Collectors.toList()); Collections.sort(stone_list); int size = stone_list.size(); while(true){ if(size == 0) return 0; else if(size == 1) return stone_list.get(0); else{ int x = stone_list.get(size-1); int y = stone_list.get(size-2); stone_list.remove(size-1); stone_list.remove(size-2); if(x != y){ stone_list.add(x-y); } Collections.sort(stone_list); size = stone_list.size(); } } } }