322. Coin Change

标签：ati   use   blog   amount   discuss   []   经典的   http   make   

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.

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

非常经典的dp问题

将amount作为dp的值，得到每个amount最少的硬币数

init ： dp[0] = 0;

regular : dp[amount] = Math.min(dp[amount], dp[amount - coins[j]] + 1); 需设dp［i］＝ max;

o(n ^2);

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

322. Coin Change

标签：ati   use   blog   amount   discuss   []   经典的   http   make   

原文地址：http://www.cnblogs.com/joannacode/p/5998866.html