LeetCode OJ 322. Coin Change DP求解

Posted 2020-09-27 yangykaifa

    题目链接：https://leetcode.com/problems/coin-change/

322. Coin Change

My Submissions

Question

Total Accepted: 15289 Total Submissions: 62250 Difficulty: Medium

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.

Subscribe to see which companies asked this question

Show Tags

Have you met this question in a real interview? 

Yes

 

No

Discuss

    给定几个固定面值的硬币，能够无限使用。

一个目标数。要求用最少的硬币兑换这个target。

    换一种思路理解题目，每次能够走给定面值的步数。问最少走多少步能达到目标。

如此一来便能够用BFS求解。

    另外一种解法是DP，dp[i] = min {dp[i - a], dp[i - b], dp[i - c] ...... }。动态规划仅仅要有公式就能非常快求解。

    我用DP求解的AC代码

package march;

import java.util.Arrays;

/**
 * @auther lvsheng
 * @date 2016年3月16日
 * @time 下午11:20:44
 * @project LeetCodeOJ
 * 
 */

public class CoinChange {
	public static void main(String[] args) {
		int[] a = { 1, 2, 5 };
		System.out.println(coinChange(a, 11));
		int[] b = { 2 };
		System.out.println(coinChange(b, 3));
	}

	public static int coinChange(int[] coins, int amount) {
		int len = coins.length;
		if (len == 0) return -1;
		int[] dp = new int[amount + 1];
		Arrays.fill(dp, Integer.MAX_VALUE);
		dp[0] = 0;

		Arrays.sort(coins);

		for (int i = 0; i < len && coins[i] <= amount; i++) dp[coins[i]] = 1;

		for (int i = 1; i <= amount; i++) {
			int min = dp[i];
			if (min == 1) continue;
			for (int j = 0; j < len && coins[j] < i; j++) {
				int c = coins[j];
				int d = dp[i - c];
				if (d != Integer.MAX_VALUE && d < min) min = d + 1;
			}
			dp[i] = min;
		}

		return dp[amount] == Integer.MAX_VALUE ? -1 : dp[amount];
	}
}