Last Stone Weight II

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You are given an array of integers stones where stones[i] is the weight of the ith stone.

We are playing a game with the stones. On each turn, we choose any two 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 destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.
At the end of the game, there is at most one stone left.

Return the smallest possible weight of the left stone. If there are no stones left, return 0.

Example1:

Input: stones = [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.

Example2:

Input: stones = [31,26,33,21,40]
Output: 5

Example3:

Input: stones = [1,2]
Output: 1

题目中文释义：

题解：

从上面题目的意思，其实我们可以转换一种角度来看待这个问题：

这个题目实质上其实就是，将所有的石子分为两堆，求两堆石子的最小绝对值的问题，最理想的答案是两堆石子一样多，但实际上可能会有偏差；假定所有的石子和为sum, 那么最终的一堆石子中必定有一堆的石子总和<=sum/2的，并且我们让其最大化。

因此，这个问题就可以转为背包为sum/2的01背包问题, 假定其值为dp[cap]
那么，最终的结果为 sum - 2*dp[cap]
证明：两堆石子的重量分别为x,y
结果为：|x-y|
假设x为其中较小的，那么结果为: y-x = (x+y)-2*x

因此，这问题就出来了，如下：

	
	public int lastStoneWeightII(int[] stones) {
        int sum = 0;
        for (int stone : stones) {
            sum += stone;
        }

        int cap = sum / 2;
        int[] dp = new int[cap +1];
        for (int stone : stones) {
            for (int i = cap; i >= stone; i--) {
                dp[i] = Math.max(dp[i],dp[i-stone]+stone);
            }
        }

        return sum - 2*dp[cap];

    }