题目
LintCode92.背包问题
在n个物品中挑选若干物品装入背包,最多能装多满?假设背包的大小为m,每个物品的大小为A[i]
样例
样例 1:
输入: [3,4,8,5], backpack size=10
输出: 9
样例 2:
输入: [2,3,5,7], backpack size=12
输出: 12
挑战
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.
注意事项
你不可以将物品进行切割。
题解
采用动态规划实现
dp[i][j] = dp[i-1][j-A[i]]+A[i] > dp[i-1][j]?dp[i-1][j-A[i]]+A[i]:dp[i-1][j];
C++实现
#include <iostream>
#include <vector>
using namespace std;
int backPack(int m, vector<int> &A);
int main(int argc, const char * argv[]) {
vector<int> A = {3,4,8,5};
cout<<"result = "<<backPack(10, A)<<endl;
}
int backPack(int m, vector<int> &A) {
vector<vector<int>> dp(A.size(),vector<int>(m+1,0));
int size = A.size();
for(int i = 0 ;i<size;i++){
for(int j=0;j<m+1;j++){
if(i==0){
if(A[i]<=j)
dp[i][j] = A[i];
}else if(A[i]<=j){
dp[i][j] = dp[i-1][j-A[i]]+A[i] > dp[i-1][j]?dp[i-1][j-A[i]]+A[i]:dp[i-1][j];
}else{
dp[i][j] = dp[i-1][j];
}
}
}
return dp[size-1][m];
}