[Leetcode] 322. Coin Change 解题报告

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.

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

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

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

思路：

一道难度适中的动态规划题目。定义dp[i]表示要凑够i块钱需要的最小硬币数量，则状态转移方程为：dp[i] = min(dp[i - coins[j]] + 1), 0 <= j < coins.size()。在实现的时候需要注意两个合法条件：1）i - coins[j] 必须大于等于0，并且dp[0]要初始化为0；2）dp[i - coins[j]]必须是已经被计算过的合法数。该算法的时间复杂度是O(amount * coins.size())，空间复杂度是O(amount)。

代码：

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