【LeetCode】322. Coin Change

Difficulty: Medium

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.

f[i]表示组成和为i的最少数;

f[0]=0;

状态转移方程f[i]=min{ f[ i-nums[j] ] }+1;

算法时间复杂度为O(amount * coins.size() )

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

return f[amount];
}
};