dp/knapsack-problem/last-stone-weight-ii

Last Stone Weight II

描述

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.

Note:

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

分析

两块石头粉碎的过程，就是给每个石头赋值正号或者负号的过程，相当于把这些石头分为两组，使得两组的差值尽可能小，所以这是经典的集合划分 NP 完全问题，可以采用动态规划的方法求解。

代码

方法 1

可以转换成 0-1 背包问题，更加简化，只有重量没有价值信息。每个物品 i 的重量为 stones[i]，价值为 0，背包能容纳的最大重量为sum(stones)/2。该问题就变成，选择若干物品，能否恰好填满背包？令 f[i][j]表示前 i 个物品能否填满容量为 j 的背包，则状态转移方程为:

$$f[i][j] = f[i-1][j] \lor f[i-1][j-W[i]]$$

# TODO

// Last Stone Weight II
// 0-1 knapsack problem
// Time Complexity: O(n*W), Space Complexity: O(W)
class Solution {
    public int lastStoneWeightII(int[] stones) {
        int sum = 0;
        for(int i : stones) sum += i;

        int[] w = stones; // weight array
        int W = sum / 2; // maximum weight capacity of knapsack

        boolean[] f = new boolean[W + 1];
        f[0] = true; // initialize

        for(int i = 0; i < stones.length; i++) {
            for(int j = W; j >= w[i]; --j) {
                f[j] = f[j] || f[j-w[i]];
            }
        }

        for (int j = W; j >= 0; j--)
            if (f[j]) return sum - j*2;
        return -1;
    }
}

// TODO

方法 2

可以转换成标准的 0-1 背包问题，每个物品 i 的重量为 stones[i]，价值也为stones[i]，背包能容纳的最大重量为sum(stones)/2。

# TODO

// Last Stone Weight II
// 0-1 knapsack problem
// Time Complexity: O(n*W), Space Complexity: O(W)
class Solution {
    public int lastStoneWeightII(int[] stones) {
        int sum = 0;
        for(int i : stones) sum += i;

        int[] w = stones; // weight array
        int W = sum / 2; // maximum weight capacity of knapsack

        int[] f = new int[W + 1];
        f[0] = 0; // initialize

        for(int i = 0; i < stones.length; i++) {
            for(int j = W; j >= w[i]; --j) {
                f[j] = Math.max(f[j], f[j-w[i]]+w[i]);
            }
        }

        return sum - f[W]*2;
    }
}

// TODO

相关题目

• Partition Equal Subset Sum