Lintcode - Backpack II

Given n items with size A[i] and value V[i], and a backpack with size m. What's the maximum value can you put into the backpack?

Note

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

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.

思路：

參見 to http://www.geeksforgeeks.org/dynamic-programming-set-10-0-1-knapsack-problem/

這道題很類似於word cut，都是要掃描數組，並且對於每一個數組再進行一次for loop尋找對於該元素的最大值。

對於每一個數組元素我們可以選擇加入或者不加入揹包。如果元素重量小於等於當前目標重量，那麼可以選擇加入揹包；如果大於目標重量，那麼該元素無法加入，揹包的價值等於上一個元素的最大值。

d[i][j] = max(d[i-1][j], d[i-1][j-A[i-1]] + V[i-1]) (given A[i-1]<=j) or d[i-1][j] (given A[i-1]>j)

    public int backPackII(int m, int[] A, int V[]) {
        // write your code here
        int[][] d = new int[A.length+1][m+1];
        
        for (int i = 0; i <= A.length; i++) {
            for (int j = 0; j <= m; j++) {
                if (i == 0 || j == 0) {
                    d[i][j] = 0;
                } else {
                    if (A[i-1] <= j) {
                        d[i][j] = Math.max(d[i-1][j], d[i-1][j-A[i-1]] + V[i-1]);
                    } else {
                        d[i][j] = d[i-1][j];
                    }
                }
            }
        }
        return d[A.length][m];
    }

Use one dimension array:

    public int backPackII(int m, int[] A, int V[]) {
        int[] d = new int[m+1];

        for (int i = 1; i <= A.length; i++) {
            for (int j = m; j >= 0; j--) {
                if (j == 0) {
                    d[j] = 0;
                } else {
                    if (A[i-1] <= j) {
                        d[j] = Math.max(d[j-A[i-1]]+V[i-1], d[j]);
                    }
                }
            }
        }
        return d[m];
    }

O(nm)