LeetCode 1049. Last Stone Weight II

Description:
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 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 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.
Solution:
思路如下：经典0-1背包问题，由于码代码的时候懒得切换输入法，大家将就着看～

//choose one set of stones whose sum is mostly half of the sum of all numbers
//if two sets are divided into 2 almost the same parts, then their difference will be the least
//then the problem can be thought as picking stones into a bag with size == sum / 2;
//now, you will see it as a so-called “0-1 bag issue”

public int lastStoneWeightII(int[] stones) {
//choose one set of stones whose sum is mostly half of the sum of all numbers
//if two sets are divided into 2 almost the same parts, then their difference will be the least
//then the problem can be thought as picking stones into a bag with size == sum / 2;
//now, you will see it as a so-called "0-1 bag issue"
int sum = 0;
for(int i : stones)
sum += i;
int subSum = sum / 2;
int[][] dp = new int[stones.length + 1][subSum + 1];
for(int i = 1; i <= stones.length; ++i) {
for(int j = 1; j <= subSum; ++j) {
if(stones[i - 1] > j)
dp[i][j] = dp[i - 1][j];
else //the bag can contain the value of the stone
dp[i][j] = Math.max(dp[i - 1][j], dp[i - 1][j - stones[i - 1]] + stones[i - 1]);
}
}
subSum = dp[stones.length][subSum];
return Math.abs(sum - 2 * subSum);
}