leetcode 322. Coin Change

题目描述：

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.

很简单的动态规划，其中dp[i]表示和为i所需要的最少的硬币数，

ac代码：

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

另附上leetcode上的题解，上面有各种解法的比较和分析（暴力解法，自顶向下的dp和自底向上的dp）