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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ii的重量wiw_istones[i],价值viv_i也为stones[i],背包能容纳的最大重量W=12iwiW=\frac{1}{2}\sum_i w_i

代码

// 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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