Backpack

Posted 2021-03-21 flagyuri

Description

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

Example 1:
	Input:  [3,4,8,5], backpack size=10
	Output:  9

Example 2:
	Input:  [2,3,5,7], backpack size=12
	Output:  12
	

Challenge

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];
    }
}