标签：解法 logs bsp amount example any strong pack length
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.
和之前的BackPackIV解法雷同
1 public class Solution { 2 public int coinChange(int[] coins, int amount) { 3 if(amount<=0) return 0; 4 5 int[] t = new int[amount+1]; 6 t[0] =0; 7 for(int i=1;i<=amount;i++){ 8 t[i] = Integer.MAX_VALUE; 9 } 10 for(int i=1;i<=amount;i++){ 11 for(int j=0;j<coins.length;j++){ 12 if(coins[j]<=i){ 13 int pre = t[i-coins[j]]; 14 if(pre!=Integer.MAX_VALUE) 15 t[i] = Math.min(t[i], pre+1); 16 } 17 } 18 } 19 return t[amount]==Integer.MAX_VALUE?-1:t[amount]; 20 } 21 }
标签：解法 logs bsp amount example any strong pack length
原文地址：http://www.cnblogs.com/xinqiwm2010/p/6836077.html
