Backpack | & ||

Backpack |

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

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.

分析：

看似这题是NP-hard问题，但是实际上可以用DP解决。result[i][j] 表示选取数组A中前i个数并且backpack size 是 j的时候，backpack剩余的size最小。

result[i][j] = Math.min(result[i - 1][j], result[i - 1][j - A[i]]);

 public class Solution {

     public int backPack(int m, int[] A) {
         if (A == null || A.length ==  || m <= ) return m;

         int[][] result = new int[A.length][m + ];
         for (int i = ; i < result.length; i++) {
             for (int j = ; j <= m; j++) {
                 if (i == ) {
                     if (A[i] > j) {
                         result[i][j] = j;
                     } else {
                         result[i][j] = j - A[i];
                     }
                 } else {
                     if (A[i] > j) {
                         result[i][j] = result[i - ][j];
                     } else {
                         result[i][j] = Math.min(result[i - ][j], result[i - ][j - A[i]]);
                     }
                 }

             }
         }
         return m - result[A.length - ][m];
     }
 }

Backpack II

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?

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.

分析：

原理同上，转移方程如下：

maxValue[i][j] = Math.max(maxValue[i - 1][j], maxValue[i - 1][j - A[i]] + V[i]);

 public class Solution {
     public int backPackII(int m, int[] A, int V[]) {
         if (m <=  || A == null || A.length ==  || V == null || V.length == ) return ;

         int[][] maxValue = new int[A.length][m + ];

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

参考请注明出处：cnblogs.com/beiyeqingteng/