322. Coin Change

322. Coin Change

题目：

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

Example 1:

coins = [1, 2, 5], amount = 11return 3 (11 = 5 + 5 + 1)

Example 2:

coins = [2], amount = 3return -1.

代码：

暴力解法（超时）：

class Solution {public:    int coinChange(vector<int>& coins, int amount) {        sort(coins.begin(), coins.end());        vector<vector<int>> answer;        vector<int> sol;        getCom(sol, answer, coins, amount, 0);        if (answer.size() == 0) return -1;        else {            vector<int> min = answer[0];            for (int i = 0; i < answer.size(); i++) {                if (answer[i].size() < min.size())                    min = answer[i];            }            return min.size();        }    }    void getCom(vector<int>& sol, vector<vector<int>>& res, vector<int>& coins, int amount, int start) {        if (!amount) {            res.push_back(sol);            return;        }        for (int i = start; i < coins.size() && amount >= coins[i]; i++) {            sol.push_back(coins[i]);            getCom(sol, res, coins, amount - coins[i], i);            sol.pop_back();        }    }};

优化版：

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