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 weightx
is totally destroyed, and the stone of weighty
has new weighty-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,令每个物品的重量为 stones[i]
,价值也为stones[i]
,背包能容纳的最大重量。
代码
- Python
- Java
- C++
# 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[] 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;
}
}
// 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;
}
};