Leetcode: 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:
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.

类似于经典的DP Coin change求最多的组合次数，这里求最小的硬币总数。采用由底往上的解法，从0开始。

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