|
1 | 1 | packagecom.fishercoder.solutions; |
2 | 2 |
|
3 | | -/** |
4 | | - * 1049. Last Stone Weight II |
5 | | - * |
6 | | - * We have a collection of rocks, each rock has a positive integer weight. |
7 | | - * Each turn, we choose any two rocks and smash them together. |
8 | | - * Suppose the stones have weights x and y with x <= y. The result of this smash is: |
9 | | - * If x == y, both stones are totally destroyed; |
10 | | - * If x != y, the stone of weight x is totally destroyed, and the stone of weight y has new weight y-x. |
11 | | - * At the end, there is at most 1 stone left. Return the smallest possible weight of this stone (the weight is 0 if there are no stones left.) |
12 | | - * |
13 | | - * Example 1: |
14 | | - * Input: [2,7,4,1,8,1] |
15 | | - * Output: 1 |
16 | | - * |
17 | | - * Explanation: |
18 | | - * We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then, |
19 | | - * we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then, |
20 | | - * we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then, |
21 | | - * we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value. |
22 | | - * |
23 | | - * Note: |
24 | | - * 1 <= stones.length <= 30 |
25 | | - * 1 <= stones[i] <= 100 |
26 | | - * */ |
27 | 3 | publicclass_1049 { |
28 | 4 | publicstaticclassSolution1 { |
29 | 5 | publicintlastStoneWeightII(int[]stones) { |
|