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：
要求满足amount时用的最少coins数目.
Let dp[i] denotes amount == i 时用的最少coins数目. 用一维array储存.
递推时，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].
初始化dp[0] = 0. amount为0时不需要硬币. dp其他值都是Integer的最大值, 之后会每次更新取最小值.
Time Complexity: O(amount * coins.length).
Space: O(amount).
AC Java:
class Solution {
public int coinChange(int[] coins, int amount) {
if(amount == 0){
return 0;
}
if(coins == null || coins.length == 0){
return -1;
}
int [] dp = new int[amount+1];
dp[0] = 0;
for(int i = 1; i<=amount; i++){
dp[i] = Integer.MAX_VALUE;
for(int coin : coins){
if(i-coin<0 || dp[i-coin]==Integer.MAX_VALUE){
continue;
}else{
dp[i] = Math.min(dp[i], dp[i-coin]+1);
}
}
}
return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
}
}