标签：items cep show pac time 思路 into @param end
Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?
You can not divide any item into small pieces.
Example 1:
Input: [3,4,8,5], backpack size=10
Output: 9
Example 2:
Input: [2,3,5,7], backpack size=12
Output: 12
O(n x m) time and O(m) memory.
O(n x m) memory is also acceptable if you do not know how to optimize memory.
思路：f[i][j]表示前i个物品选一些物品放入容量为j的背包中能否放满。
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
public int backPack(int m, int[] A) {
boolean f[][] = new boolean[A.length + 1][m + 1];
for (int i = 0; i <= A.length; i++) {
for (int j = 0; j <= m; j++) {
f[i][j] = false;
}
}
f[0][0] = true;
for (int i = 1; i <= A.length; i++) {
for (int j = 0; j <= m; j++) {
f[i][j] = f[i - 1][j];
if (j >= A[i-1] && f[i-1][j - A[i-1]]) {
f[i][j] = true;
}
} // for j
} // for i
for (int i = m; i >= 0; i--) {
if (f[A.length][i]) {
return i;
}
}
return 0;
}
}
O(m) 空间复杂度的解法
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
public int backPack(int m, int[] A) {
int f[] = new int[m + 1];
for (int i = 0; i < A.length; i++) {
for (int j = m; j >= A[i]; j--) {
f[j] = Math.max(f[j], f[j - A[i]] + A[i]);
}
}
return f[m];
}
}
标签：items cep show pac time 思路 into @param end
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078365.html
