1049. 最後一塊石頭的重量 II
class Solution {
public int lastStoneWeightII(int[] stones) {
int sum = 0;
for(int i = 0;i < stones.length;i++){
sum += stones[i];
}
int target = sum / 2;
int[][] dp = new int[stones.length + 1][target + 1];
for(int i = 1; i <= stones.length;i++){
int x = stones[i - 1];
for(int j = 0; j <= target;j++){
dp[i][j] = dp[i - 1][j];
if(j >= x) dp[i][j] = Math.max(dp[i][j],dp[i - 1][j - x] + x);
}
}
int res = sum - 2*dp[stones.length][target];
return res;
}
}
class Solution {
public int lastStoneWeightII(int[] stones) {
int sum = 0;
for(int i = 0;i < stones.length;i++){
sum += stones[i];
}
int target = sum / 2;
int[] dp = new int[target + 1];
for(int i = 1; i <= stones.length;i++){
int x = stones[i - 1];
for(int j = target; j >= x;j--){
dp[j] = Math.max(dp[j],dp[j - x] + x);
}
}
int res = sum - 2*dp[target];
return res;
}
}
思路還是可以轉變為0-1背包問題,重量一半的情況下,能裝下的最大值,使左右兩邊的石頭盡量均衡,這樣可以達到碰撞後重量最小。