动态规划------Backpack II

最新推荐文章于 2023-06-07 23:56:49 发布

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本文深入探讨了背包问题的优化算法，从传统的O(nxm)时间复杂度改进到O(m)，并通过具体示例阐述了实现方法及关键步骤。重点在于如何在内存使用上进行优化，使解决方案更适用于资源受限的场景。

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Given n items with size Ai and value Vi, and a backpack with size m. What’s the maximum value can you put into the backpack?

Have you met this question in a real interview? Yes
Example
Given 4 items with size [2, 3, 5, 7] and value [1, 5, 2, 4], and a backpack with size 10. The maximum value is 9.

Note
You cannot divide item into small pieces and the total size of items you choose should smaller or equal to m.

Challenge
O(n x m) memory is acceptable, can you do it in O(m) memory?

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public class Solution {
    /**
     * @param m: An integer m denotes the size of a backpack
     * @param A & V: Given n items with size A[i] and value V[i]
     * @return: The maximum value
     */
    public int backPackII(int m, int[] A, int V[]) {
        int[][] res = new int[A.length+1][m+1];
        res[0][0] = 0;
        for (int i=1; i<=A.length; i++) {
            for (int j=0; j<=m; j++) {
                if (j - A[i-1] < 0)
                    res[i][j] = res[i-1][j];
                if (j - A[i-1] >= 0) {
                    res[i][j] = Math.max(res[i-1][j], res[i-1][j-A[i-1]]+V[i-1]);
                }
            }
        }

        return res[A.length][m];
    }
}