原题链接在这里：https://leetcode.com/problems/last-stone-weight-ii/
题目：
We have a collection of rocks, each rock has a positive integer weight.
Each turn, we choose any two 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 weightxis totally destroyed, and the stone of weightyhas new weighty-x.
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.)
Example 1:
Input: [2,7,4,1,8,1] Output: 1 Explanation: We can combine 2 and 4 to get 2 so the array converts to [2,7,1,8,1] then, we can combine 7 and 8 to get 1 so the array converts to [2,1,1,1] then, we can combine 2 and 1 to get 1 so the array converts to [1,1,1] then, we can combine 1 and 1 to get 0 so the array converts to [1] then that's the optimal value.
Note:
1 <= stones.length <= 301 <= stones[i] <= 100
题解：
If we choose any two rocks, we could divide the rocks into 2 groups.
And calculate the minimum diff between 2 groups' total weight.
Since stones.length <= 30, stone weight <= 100, maximum total weight could be 3000.
Let dp[i] denotes whether or not smaller group weight could be i. smaller goupe total weight is limited to 1500. Thus dp size is 1501. It becomes knapsack problem.
dp[0] = true. We don't need to choose any stone, and we could get group weight as 0.
For each stone, if we choose it we track dp[i-w] in the previoius iteration. If we don't choose it, track dp[i] from last iteration. Thus it is iterating from big to small.
Time Complexity: O(n*sum). n = stones.length. sum is total weight of stones.
Space: O(sum).
AC Java:
1 class Solution {
2 public int lastStoneWeightII(int[] stones) {
3 if(stones == null || stones.length == 0){
4 return 0;
5 }
6
7 int sum = 0;
8 boolean [] dp = new boolean[1501];
9 dp[0] = true;
10
11 for(int w : stones){
12 sum += w;
13
14 for(int i = Math.min(sum, 1501); i>=w; i--){
15 dp[i] = dp[i] | dp[i-w];
16 }
17 }
18
19 for(int i = sum/2; i>=0; i--){
20 if(dp[i]){
21 return sum-i-i;
22 }
23 }
24
25 return 0;
26 }
27 }