LeetCode 1049. Last Stone Weight II

Posted 2020-11-21 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/last-stone-weight-ii/

题目：

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

题解：

If we choose any two rocks, we could divide the rocks into 2 groups.

And calculate the minimum diff between 2 groups\' total weight.

Since stones.length <= 30, stone weight <= 100, maximum total weight could be 3000.

Let dp[i] denotes whether or not smaller group weight could be i. smaller goupe total weight is limited to 1500. Thus dp size is 1501. It becomes knapsack problem.

dp[0] = true. We don\'t need to choose any stone, and we could get group weight as 0.

For each stone, if we choose it we track dp[i-w] in the previoius iteration. If we don\'t choose it, track dp[i] from last iteration. Thus it is iterating from big to small.

Time Complexity: O(n*sum). n = stones.length. sum is total weight of stones.

Space: O(sum).

AC Java: 

 1 class Solution {
 2     public int lastStoneWeightII(int[] stones) {
 3         if(stones == null || stones.length == 0){
 4             return 0;
 5         }
 6         
 7         int sum = 0;
 8         boolean [] dp = new boolean[1501];
 9         dp[0] = true;
10         
11         for(int w : stones){
12             sum += w;
13             
14             for(int i = Math.min(sum, 1501); i>=w; i--){
15                 dp[i] = dp[i] | dp[i-w];
16             }
17         }
18         
19         for(int i = sum/2; i>=0; i--){
20             if(dp[i]){
21                 return sum-i-i;
22             }
23         }
24         
25         return 0;
26     }
27 }

是Last Stone Weight进阶.