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

本题可以转换成标准的 0-1 背包问题，令每个物品$i$的重量$w_i$为 stones[i]，价值$v_i$也为stones[i]，背包能容纳的最大重量$W=\frac{1}{2}\sum_i w_i$。

代码

Python

# TODO

Java

// 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[] dp = new int[W + 1];
        dp[0] = 0; // base case

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

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

C++

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

        const vector<int>& w = stones; // weight array
        int W = sum / 2; // maximum weight capacity of knapsack

        vector<int> dp(W + 1);
        dp[0] = 0; // base case

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

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

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