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322. Coin Change

LeetCode - Coin Change

0322 01

思考题：

  1. 如何把动态规划代码写得更简洁一些？

  2. 如何使用自底向上方法来实现一遍动态规划？

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:

Input: coins = [1, 2, 5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Note:

You may assume that you have an infinite number of each kind of coin.

思路分析

解题思路

  1. 这种找路径，找方法的题一般可以使用回溯法来解决，回溯法也可以说是树形图法，解题的时候使用类似于树状图的结构，使用 自顶而下 的方法。

  2. 而在回溯法中，如果含有很多的重复的计算的时候，就可以使用记忆化的搜索，将可能出现的重复计算大状态使用一个数组来保存其值，在进行重复的计算的时候，就可以直接的调用数组中的值，较少了不必要的递归。

  3. 使用了记忆化搜索后，一般还可以进行优化，在记忆化搜索的基础上，变成 自底而上 的动态规划。

0322 03

暴力穷举

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/**
 * 最笨的方法：暴力穷举，超时
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2024-06-27 21:12
 */
public int coinChange(int[] coins, int amount) {
  if (amount < 0) {
    return -1;
  }
  if (amount == 0) {
    return 0;
  }
  int result = Integer.MAX_VALUE;
  for (int coin : coins) {
    int cnt = coinChange(coins, amount - coin);
    if (cnt == -1) {
      continue;
    }
    result = Math.min(result, cnt + 1);
  }
  return result == Integer.MAX_VALUE ? -1 : result;
}

动态规划

0322 02 
F(3) =min(F(3−c1),F(3−c 2),F(3−c3))+1
     =min(F(3−1),F(3−2),F(3−3))+1
     =min(F(2),F(1),F(0))+1
     =min(1,1,0)+1
     =1
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/**
 * 参考 https://leetcode.cn/problems/coin-change/solutions/132979/322-ling-qian-dui-huan-by-leetcode-solution/[322. 零钱兑换 - 官方题解^]
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2024-06-27 21:12
 */
public int coinChange(int[] coins, int amt) {
  if (coins == null || coins.length == 0 || amt < 0) {
    return -1;
  }
  int max = amt + 1;
  int[] dp = new int[max];
  Arrays.fill(dp, max);
  dp[0] = 0;
  for (int i = 1; i <= amt; i++) {
    for (int coin : coins) {
      if (coin <= i) {
        dp[i] = Math.min(dp[i], dp[i - coin] + 1);
      }
    }
  }
  return dp[amt] > amt ? -1 : dp[amt];
}

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/**
 * Runtime: 91 ms, faster than 6.13% of Java online submissions for Coin Change.
 *
 * Memory Usage: 42.7 MB, less than 5.33% of Java online submissions for Coin Change.
 *
 * Copy from: https://leetcode.com/problems/coin-change/solution/[Coin Change solution - LeetCode]
 */
public int coinChange(int[] coins, int amount) {
    return coinChange(coins, amount, new int[amount]);
}

private int coinChange(int[] coins, int current, int[] count) {
    if (current < 0) {
        return -1;
    }
    if (current == 0) {
        return 0;
    }
    if (count[current - 1] != 0) {
        return count[current - 1];
    }
    int min = Integer.MAX_VALUE;
    for (int coin : coins) {
        int res = coinChange(coins, current - coin, count);
        if (0 <= res && res < min) {
            min = res + 1;
        }
    }
    count[current - 1] = (min == Integer.MAX_VALUE) ? -1 : min;
    return count[current - 1];
}

/**
 * Runtime: 427 ms, faster than 5.00% of Java online submissions for Coin Change.
 *
 * Memory Usage: 115.5 MB, less than 5.33% of Java online submissions for Coin Change.
 */
public int coinChangeDpWithMemo(int[] coins, int amount) {
    int count = dpWithMemo(coins, new HashMap<>(), amount, amount);
    if (count > amount) {
        return -1;
    }
    return count;
}

public int dpWithMemo(int[] coins, Map<Integer, Integer> memo, int amount, int current) {
    if (current == 0) {
        return 0;
    }
    if (current < 0) {
        return amount;
    }

    List<Integer> counts = new ArrayList<>();
    for (int coin : coins) {
        int newCoin = current - coin;
        int count;
        if (memo.containsKey(newCoin)) {
            count = memo.get(newCoin);
        } else {
            count = dpWithMemo(coins, memo, amount, newCoin) + 1;
            memo.put(newCoin, count);
        }
        counts.add(count);
    }
    return Collections.min(counts);
}

/**
 * Time Limit Exceeded
 */
public int coinChangeDp(int[] coins, int amount) {
    int count = dp(coins, amount, amount);
    if (count > amount) {
        return -1;
    }
    return count;
}

private int dp(int[] coins, int amount, int current) {
    if (current == 0) {
        return 0;
    }
    if (current < 0) {
        return amount;
    }

    List<Integer> counts = new ArrayList<>();
    for (int coin : coins) {
        counts.add(dp(coins, amount, current - coin) + 1);
    }
    return Collections.min(counts);
}

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/**
 * 参考 左程云《程序员代码面试指南》。完全没看明白。
 *
 * @author D瓜哥 · https://www.diguage.com
 * @since 2024-06-27 21:12
 */
public int coinChange(int[] coins, int amt) {
  if (coins == null || coins.length == 0 || amt < 0) {
    return -1;
  }
  int N = coins.length;
  int[][] dp = new int[N + 1][amt + 1];
  // 设置最后一排的值，除了 dp[N][0] 外，其他都是-1
  for (int i = 1; i <= amt; i++) {
    dp[N][i] = -1;
  }
  for (int i = N - 1; i >= 0; i--) {  // 从底向上计算每一行
    printMatrix(dp);
    for (int rest = 0; rest <= amt; rest++) { // 每一行都从左向右
      dp[i][rest] = -1; // 初始时，先设置 dp[i][rest] 的值无效。这样省去了一次遍历设置每一行的操作
      if (dp[i + 1][rest] != -1) { // 下面的值如果有效
        dp[i][rest] = dp[i + 1][rest]; // 先设置成下面的值
      }
      // 如果坐标的位置不越界且有效
      if (rest - coins[i] >= 0 && dp[i][rest - coins[i]] != -1) {
        if (dp[i][rest] == -1) { // 如果之前下面的值无效
          dp[i][rest] = dp[i][rest - coins[i]] + 1;
        } else { // 说明下面和左边的值都有效，取最小的
          dp[i][rest] = Math.min(dp[i][rest], dp[i][rest - coins[i]] + 1);
        }
      }
    }
  }
  System.out.println("最后结果");
  printMatrix(dp);
  return dp[0][amt];
}

参考资料

  1. 322. 零钱兑换 - 官方题解

  2. 322. 零钱兑换 - Java 递归、记忆化搜索、动态规划

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