leetcode 1049

问题

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:

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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. 1 <= stones.length <= 30
  2. 1 <= stones[i] <= 100

分析

该问题可以转化为“给定一堆石头，分成两个部分，要求两个部分之差最小，求这个最小值。”，那么我们可以使用背包模型来解这个问题。假设所有石头的总重量是 sum，那么现在有一个容量为 halfSum = sum/2 的背包，每个是石块的重量与其体积相等，现在向背包中装石头，问最多能容纳的重量是多少？假设为 x ，那么这道题的答案就是 sum-2*x。

https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/1049.Last-Stone-Weight-II

代码

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class Solution {
public:
    int lastStoneWeightII(vector<int>& stones) {
        int sum = accumulate(stones.begin(), stones.end(), 0);
        int halfSum = sum/2;
        bitset<1501> dp;
        dp[0] = 1;
        for (int a: stones) {
            for (int i = halfSum; i >= 0; --i) {
                if (i-a >= 0)
                    dp[i] = dp[i] | dp[i-a];
            }
        }
        for (int i = halfSum; i >= 0; --i) {
            if (dp[i] == 1) {
                return (sum-i) - i;
            }
        }
        return 0;
    }
};