LeetCode week 15 ： Coin Change2

题目

地址： https://leetcode.com/problems/coin-change-2/description/
类别： 动态规划
难度： Medium
描述：

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.

Example 1:

Input: amount = 5, coins = [1, 2, 5]Output: 4Explanation: there are four ways to make up the amount:5=55=2+2+15=2+1+1+15=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]Output: 0Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

Input: amount = 10, coins = [10] Output: 1

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

分析

给定含n种面值不同的硬币，每种硬币皆有无限多个，判断能否用这些硬币凑出总金额amount，若不能，则返回0；若能，则返回组合总数。
例：

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

共有以下4种组合：

5=55=2+2+15=2+1+1+15=1+1+1+1+1

因此：

Output: 4

思路

动态规划

定义子问题dp[i][j] : 用前i种硬币组成金额j的总组合数
最终目标：dp[n][amount]
状态迁移方程：
求dp[i][j]时：

 1. 不用第i种硬币，（1）用不了:j<coins[i]; （2）能用但不用。 两种情况都只用前 i-1 种硬币去组成j， 此时共有dp[i-1][j]种组合； 2. 用第i种硬币, 此时因为硬币无限，所以它的前一个状态仍能用i种硬币，即此时共有dp[i][j-coins[i]]种组合。因此总的组合数是两种情况的和：dp[i][j] = dp[i-1][j] + dp[i][j-coins[i-1]]

代码：

class Solution {public:    int change(int amount, vector<int>& coins) {        vector<vector<int>> dp(coins.size()+1, vector<int>(amount+1));        dp[0][0] = 1;        for (int i = 1; i <= coins.size(); i++) {            dp[i][0] = 1;            for (int j = 1; j <= amount; j++) {                dp[i][j] = dp[i-1][j] + (j >= coins[i-1] ? dp[i][j-coins[i-1]] : 0);            }        }        return dp[coins.size()][amount];    }};

因为dp[i][j] 只依赖于dp[i-1][j] 和 dp[i][j-coins[i]]，所以我们可以用一维数组优化空间如下：

class Solution {public:    int change(int amount, vector<int>& coins) {        vector<int> dp(amount+1);        dp[0] = 1;        for(auto c : coins) {            for(int i = c; i <= amount; i++)                dp[i] = dp[i] + dp[i-c];        }        return dp[amount];    }};