原题链接在这里：https://leetcode.com/problems/coin-change/
题目：
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.
题解：
DP问题. 状态转移dp[i] = Math.min(dp[i], dp[i-coins[j]]+1).
dp[i-coins[j]]+1表示用最少个数coin表示i-coins[j].
Time Complexity: O(amount * coins.length). Space: O(amount).
AC Java:
1 public class Solution { 2 public int coinChange(int[] coins, int amount) { 3 if(amount == 0){ 4 return 0; 5 } 6 if(coins == null || coins.length == 0 || amount < 0){ 7 return -1; 8 } 9 int [] dp = new int[amount+1]; 10 Arrays.fill(dp, Integer.MAX_VALUE); 11 dp[0] = 0; 12 for(int i = 1; i<=amount; i++){ 13 for(int j = 0; j<coins.length; j++){ 14 //i-coins[j]越界 或者 dp[i-coins[j]]为最大值表示i-coins[j]无法用coin表示 15 if(i-coins[j] < 0 || dp[i-coins[j]] == Integer.MAX_VALUE){ 16 continue; 17 } 18 dp[i] = Math.min(dp[i], dp[i-coins[j]]+1); 19 } 20 } 21 return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount]; 22 } 23 }
原文：http://www.cnblogs.com/Dylan-Java-NYC/p/5349625.html