[LeetCode]322. Coin Change

Description:

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.

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Solution:

题意：凑金钱面额。

思路：动态规划。这里的动态规划数组不是基于面值数量的数组，而是基于需要凑整的面额数组。

class Solution {public:    int coinChange(vector<int>& coins, int amount) {        vector<int> dp(amount + 1, amount + 1);        dp[0] = 0;        for (int i = 1; i <= amount; i++) {            for (vector<int>::iterator j = coins.begin(); j != coins.end(); j++) {                if (*j <= i) {                    // 如果无法组成amount，dp[amount] = min(amount + 1, amount + 1 + 1) = amount + 1                    dp[i] = min(dp[i], dp[i - *j] + 1);                }            }        }        return dp[amount] > amount ? -1 : dp[amount];    }};