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# 1049.Last-Stone-Weight-II

## 题目描述

You are given an array of integers stones where stones[i] is the weight of the ith stone.
We are playing a game with the stones. On each turn, we choose any two stones 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 destroyed, and
If x != y, the stone of weight x is destroyed, and the stone of weight y has new weight y - x.
At the end of the game, there is at most one stone left.
Return the smallest possible weight of the left stone. If there are no stones left, return 0.
Example 1:
Input: stones = [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 , then that's the optimal value.
Example 2:
Input: stones = [31,26,33,21,40]
Output: 5
Example 3:
Input: stones = [1,2]
Output: 1
Constraints:
1 <= stones.length <= 30
1 <= stones[i] <= 100

## 代码

### Approach #1

Time: O(1) && Space: O(1)
class Solution {
public int lastStoneWeightII(int[] stones) {
int n = stones.length;
int sum = 0;
for (int i = 0; i < n; i++) {
sum += stones[i];
}
int target = sum / 2;
int[] dp = new int[target + 1];
for (int i = 0; i < n; i++) {
int val = stones[i];
for (int j = target; j >= val; j--) {
dp[j] = Math.max(dp[j], dp[j - val] + val);
}
}
return sum - 2 * dp[target];
}
}