Backpack

Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?

 Notice

You can not divide any item into small pieces.

Have you met this question in a real interview? Yes
Example
If we have 4 items with size [2, 3, 5, 7], the backpack size is 11, we can select [2, 3, 5], so that the max size we can fill this backpack is 10. If the backpack size is 12. we can select [2, 3, 7] so that we can fulfill the backpack.

You function should return the max size we can fill in the given backpack.

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.

Tags
LintCode Copyright Backpack Dynamic Programming

public int backPack(int m, int[] A) {
        // write your code here
        boolean[][] f = new boolean[A.length + 1][m + 1];

        //initialize
        f[0][0] = true;
        for (int i = 1; i <= m; i++) {
            f[0][i] = false;
        }

        for (int i = 1; i <= A.length; i++) {
            f[i][0] = true;
        }

        //function
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= A.length; j++) {
                f[j][i] = f[j - 1][i];
                if (i >= A[j - 1] && f[j - 1][i - A[j - 1]]) {
                    f[j][i] = true;
                }
            }
        }

        for (int i = m; i >= 0; i--) {
                if (f[A.length][i]) {
                   return i;
                }
            }
        return 0;
    }

　用 integer 来替代　

public int backPack(int m, int[] A) {
        // write your code here
        int[][] f = new int[A.length + 1][m + 1];

        //initialize
        f[0][0] = 0;
        for (int i = 1; i <= m; i++) {
            f[0][i] = Integer.MIN_VALUE;
        }

        for (int i = 1; i <= A.length; i++) {
            f[i][0] = 0;
        }
        int ans = 0;
        //function
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= A.length; j++) {
                f[j][i] = f[j - 1][i];
                if (i >= A[j - 1] && f[j - 1][i - A[j - 1]] != Integer.MIN_VALUE) {
                    f[j][i] = Math.max(f[j][i], f[j - 1][i - A[j - 1]] + A[j - 1]);
                    ans = Math.max(ans, f[j][i]);
                }

            }
        }
        return ans;
}