leetcode 518. Coin Change 2/硬币找零 2

归纳于http://www.cnblogs.com/grandyang/p/7669088.html

原题https://leetcode.com/problems/coin-change-2/description/

518. Coin Change 2(Medium)

You are given coins of different denominations and a total amount of money. Write a function to compute the number of

combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000

1 <= coin <= 5000

the number of coins is less than 500

the answer is guaranteed to fit into signed 32-bit integer

Example 1:

Input: amount = 5, coins = [1, 2, 5]

Output: 4

Explanation: there are four ways to make up the amount:

5=5

5=2+2+1

5=2+1+1+1

5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]

Output: 0

Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10]

Output: 1

问题描述

　　给一个总钱数amount，给一个硬币种类数组，每种硬币的数量你可以想象成足够大，问：允许使用同种硬币，一个有多少种硬币组合方式刚好能凑成总钱数amount？

分析

　　比如我们有两个硬币[1,2]，钱数为5，那么钱数的5的组成方法是可以看作两部分组成，一种是由硬币1单独组成，那么仅有一种情况(1+1+1+1+1)；另一种是由1和2共同组成，说明我们的组成方法中至少需要由一个2，所以此时我们先取出一个硬币2，那么我们只要拼出钱数为3即可，这个3还是可以用硬币1和2来拼，所以就相当于求由硬币[1,2]组成的钱数为3的总组合方法数。

动态规划：dp[i][j]表示用前i个硬币组成钱数为j的不同组合方法

递归式：dp[i][j] = dp[i - 1][j] + (j >= coins[i - 1] ? dp[i][j - coins[i - 1]] : 0)；

代码

class Solution {
public:
    int change(int amount, vector<int>& coins)
    {
        if(amount==0)
            return 1;
        if(coins.empty())
            return 0;
        vector<vector<int>> dp(coins.size()+1,vector<int>(amount+1,0));
        dp[0][0]=1;
        //dp[0][j!=0]已被初始化为0了
        for(int i=1;i<=coins.size();++i)
        {
            dp[i][0]=1;
            for(int j=1;j<=amount;++j)
            {
                if(j>=coins[i-1])
                    dp[i][j]=dp[i-1][j]+dp[i][j-coins[i-1]];
                else
                    dp[i][j]=dp[i-1][j];
/*调试用
            printf("dp[%d][%d] = dp[%d-1][%d] + (%d >= coins[%d-1] ? dp[%d][%d - coins[%d-1]] : 0)", i, j, i, j, j, i, i, j, i);
            printf("= dp[%d][%d] + (%d >= %d ? dp[%d][%d - %d] : 0)", i-1, j,j,coins[i-1],i,j, coins[i - 1]);
            dp[i][j] = dp[i - 1][j] + (j >= coins[i - 1] ? dp[i][j - coins[i - 1]] : 0);
            printf("= %d\n\n", dp[i][j]);
*/
            }
        }
        return dp[coins.size()][amount];
    }
};

原文地址：https://www.cnblogs.com/Joezzz/p/9744370.html