564.Backpack VI-背包问题 VI(中等题)

背包问题 VI

  1. 题目

    给出一个无重复正整型数组,用其中数字任意组合(可重复使用),使得和等于给定的target的所有组合。

    注意事项
    不同的序列计做不同的组合。

  2. 样例

    如nums = [1, 2, 4], target = 4
    所有组合方式如下:
    [1, 1, 1, 1]
    [1, 1, 2]
    [1, 2, 1]
    [2, 1, 1]
    [2, 2]
    [4]
    return 6

  3. 题解

public class Solution {
    /**
     * @param nums an integer array and all positive numbers, no duplicates
     * @param target an integer
     * @return an integer
     */
    public int backPackVI(int[] nums, int target) {
        int[] dp = new int[target+1];
        dp[0] = 1;
        for (int i=1;i<=target;i++)
        {
            for (int j=0;j<nums.length;j++)
            {
                if (i >= nums[j])
                    dp[i] += dp[i-nums[j]];
            }
        }

        return dp[target];
    }
}

Last Update 2016.10.7

Practice 1 Date: Monday, March 18th, 2013 We highly encourage being environment friendly and trying all problems on your own. Implement exercise 2.3-7. Implement priority queue. Implement Quicksort and answer the following questions. (1) How many comparisons will Quicksort do on a list of n elements that all have the same value? (2) What are the maximum and minimum number of comparisons will Quicksort do on a list of n elements, give an instance for maximum and minimum case respectively. Give a divide and conquer algorithm for the following problem: you are given two sorted lists of size m and n, and are allowed unit time access to the ith element of each list. Give an O(lg m + lgn) time algorithm for computing the kth largest element in the union of the two lists. (For simplicity, you can assume that the elements of the two lists are distinct). Practice 2 Date: Monday, April 1st, 2013 We highly encourage being environment friendly and trying all problems on your own. Matrix-chain product. The following are some instances. Longest Common Subsequence (LCS). The following are some instances. X: xzyzzyx Y: zxyyzxz X:MAEEEVAKLEKHLMLLRQEYVKLQKKLAETEKRCALLAAQANKESSSESFISRLLAIVAD Y:MAEEEVAKLEKHLMLLRQEYVKLQKKLAETEKRCTLLAAQANKENSNESFISRLLAIVAG Longest Common Substring. The following are some instances. X: xzyzzyx Y: zxyyzxz X:MAEEEVAKLEKHLMLLRQEYVKLQKKLAETEKRCALLAAQANKESSSESFISRLLAIVAD Y:MAEEEVAKLEKHLMLLRQEYVKLQKKLAETEKRCTLLAAQANKENSNESFISRLLAIVAG Max Sum. The following is an instance. (-2,11,-4,13,-5,-2) Shortest path in multistage graphs. Find the shortest path from 0 to 15 for the following graph.   A multistage graph is a graph (1) G=(V,E) with V partitioned into K >= 2 disjoint subsets such that if (a,b) is in E, then a is in Vi , and b is in Vi+1 for some subsets in the partition; and (2) | V1 | = | VK | = 1.     Practice 3 Date: Monday, April 15th, 2013 We highly encourage being environment friendly and trying all problems on your own. Knapsack Problem. There are 5 items that have a value and weight list below, the knapsack can contain at most 100 Lbs. Solve the problem both as fractional knapsack and 0/1 knapsack. A simple scheduling problem. We are given jobs j1, j2… jn, all with known running times t1, t2… tn, respectively. We have a single processor. What is the best way to schedule these jobs in order to minimize the average completion time. Assume that it is a nonpreemptive scheduling: once a job is started, it must run to completion. The following is an instance. (j1, j2, j3, j4) : (15,8,3,10) Single-source shortest paths. The following is the adjacency matrix, vertex A is the source.  A B C D E A -1 3 B 3 2 2 C D 1 5 E -3 All-pairs shortest paths. The adjacency matrix is as same as that of problem 3.(Use Floyd or Johnson’s algorithm)     Practice 4 Date: Monday, May 8th, 2013 We highly encourage being environment friendly and trying all problems on your own. 0/1 Knapsack Problem. There are 5 items that have a value and weight list below, the knapsack can contain at most 100 Lbs. Solve the problem using back-tracking algorithm and try to draw the tree generated. Solve the 8-Queen problem using back-tracking algorithm.    
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值