Lintcode - Backpack II

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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：

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

这道题很类似于word cut，都是要扫描数组，并且对于每一个数组再进行一次for loop寻找对于该元素的最大值。

对于每一个数组元素我们可以选择加入或者不加入背包。如果元素重量小于等于当前目标重量，那么可以选择加入背包；如果大于目标重量，那么该元素无法加入，背包的价值等于上一个元素的最大值。

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)

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