322 LeetCode Coin Change

本文介绍了一种解决硬币找零问题的高效算法。该算法利用动态规划思想,给出不同面额硬币组合成特定金额所需的最少硬币数量。若无法组合,则返回-1。适用于无限数量的同种面额硬币。

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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.


	public int coinChange(int[] coins, int amount) {
		int[] dp = new int[amount + 1];
		final int INF = 0x7ffffffe;
		for (int i = 1; i <= amount; i++) dp[i] = INF;
		for (int coin : coins)
			for (int i = coin; i <= amount; i++)
				dp[i] = Math.min(dp[i], dp[i - coin] + 1);
		return dp[amount] == INF ? -1 : dp[amount];
	}

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