LintCode Backpack II(背包II)
2016-06-21 00:55
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原题网址:http://www.lintcode.com/en/problem/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?
Notice
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
The maximum value is
方法:动态规划,与背包I问题类似。
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[]) {
// write your code here
int[] values = new int[m + 1];
Arrays.fill(values, -1);
values[0] = 0;
int max = 0;
for(int i = 0; i < A.length; i++) {
for(int j = m; j >= 0; j--) {
if (values[j] != -1 && j + A[i] <= m) {
values[j + A[i]] = Math.max(values[j + A[i]], values[j] + V[i]);
max = Math.max(max, values[j + A[i]]);
}
}
}
return max;
}
}
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?
Notice
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.
方法:动态规划,与背包I问题类似。
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[]) {
// write your code here
int[] values = new int[m + 1];
Arrays.fill(values, -1);
values[0] = 0;
int max = 0;
for(int i = 0; i < A.length; i++) {
for(int j = m; j >= 0; j--) {
if (values[j] != -1 && j + A[i] <= m) {
values[j + A[i]] = Math.max(values[j + A[i]], values[j] + V[i]);
max = Math.max(max, values[j + A[i]]);
}
}
}
return max;
}
}
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