leetcode322. [DP]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.

public int coinChange(int[] coins, int amount) {          if(coins == null || coins.length == 0) {              return -1;          }          if(amount <=0) return 0;          int[] dp = new int[amount+1];          for(int i=1;i<dp.length; i++) {              dp[i] = Integer.MAX_VALUE;          }          for(int am=1;am<=amount;am++) {              for(int i=0;i<coins.length;i++) {                  if(coins[i]<=am) {                      int diff = am - coins[i];                      if(dp[diff] != Integer.MAX_VALUE) {                          dp[am] = Math.min(dp[diff] +1, dp[am]);                      }                  }                 }          }          return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];      }  

import sysMAX_=sys.maxintclass Solution(object):    def coinChange(self, coins, amount):        if not len(coins):            return -1        if amount<=0:            return 0        dp=[MAX_ for i in range(1,amount+1)]        dp.insert(0,0)        for id in range(1,amount+1):            for j in coins:                if id>=j:                    diff=id-j                    if dp[diff]!=MAX_:                        dp[id]=min(dp[diff]+1,dp[id])#     print dp        if dp[amount]==MAX_:            return -1        else:            return dp[amount]