322. Coin Change

ou 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:这是一个动态规划题，首先dp[i],i在coins中的都是1，因为只需要一张这种coin就行了

对于i-coins[j]>=0 dp[i]=Math.max(dp[i],dp[i-coins[j]]+1)

代码如下(已通过leetcode)

public class Solution {

public int coinChange(int[] coins, int amount) {

if(coins==null||coins.length==0||amount<0) return -1;

int[] dp=new int[amount+1];

for(int i=0;i<coins.length;i++)

if(coins[i]<=amount) dp[coins[i]]=1;

for(int i=1;i<amount+1;i++) {

if(dp[i]==0) dp[i]=Integer.MAX_VALUE-1;

}

for(int i=1;i<amount+1;i++) {

for(int j=0;j<coins.length;j++) {

if(i-coins[j]>=0)

dp[i]=Math.min(dp[i], 1+dp[i-coins[j]]);

}

}

if(dp[amount]==Integer.MAX_VALUE-1) return -1;

else return dp[amount];

}

}