LeetCode322. Coin Change

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322. Coin Change

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

Example 1:

Input: coins = [1, 2, 5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Note:

You may assume that you have an infinite number of each kind of coin.

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题目：通过组合给定面额的coins，确定最少数量的coin，使得其和为amount，如果无法组成amount，返回-1。每种面额的硬币个数不限。

思路1：回溯[Time Limit Exceeded]。遍历所有可能的情况，找出满足条件的所需的最少硬币个数。

工程代码下载

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> p;
        int res = INT_MAX;
        backtracking(coins, p, 0, amount, res);
        return res == INT_MAX ? -1 : res;
    }
private:
    void backtracking(vector<int>& coins, vector<int>& p, int start, int remain, int& res){
        if(remain < 0)
            return;
        else if(remain == 0){
            res = min(res, (int)p.size());
            return;
        }

        for(int i = start; i < coins.size(); ++i){
            p.push_back(coins[i]);
            backtracking(coins, p, i, remain - coins[i], res);
            p.pop_back();
        }
    }
};

思路2：动态规划，参见解析。维护一个数组dp，dp[i]表示组成和为i所需的最少的硬币个数，外层循环i从1到amount，内层循环更新不同coin对dp的影响，判断是否能够通过coin对dp[i]进行松弛。注意dp[i]初始化为amount+1。

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        //注意，dp的初值不要设置为INT_MAX, 因为这样会导致dp[i - coins[j]] + 1可能变成负数
        vector<int> dp(amount+1, amount+1);
        dp[0] = 0;
        //计算amount_ = 1,2,3....amount时所需的最少的coins的个数
        for(int i = 1; i <= amount; ++i){
            for(auto coin : coins){
                if(i >= coin)
                    dp[i] = min(dp[i-coin] + 1, dp[i]);
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};