leetcode 322. 零钱兑换 —— 暴力递归-备忘录-DP table（Java实现）

题目

leetcode 322. 零钱兑换

解法一：暴力递归

class Solution {
    public int coinChange(int[] coins, int amount) {
        if (amount == 0) return 0;
        if (amount < 0) return -1;

        int res = Integer.MAX_VALUE;
        for (int i = 0; i < coins.length; i++) {
            if (amount < coins[i]) continue;
            int sub = coinChange(coins, amount - coins[i]);
            if (sub == -1) continue;
            res = Math.min(res, 1 + sub);
        }

        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

• 时间复杂度：O(3ⁿ)
• 空间复杂度：O(n)

解法二：备忘录，存中间结果

class Solution {
    private int[] memo;
    public int coinChange(int[] coins, int amount) {
        memo = new int[amount+1];
        Arrays.fill(memo, -6);
        return dp(coins, amount);
    }

    private int dp(int[] coins, int amount) {
        if (amount == 0) return 0;
        if (amount < 0) return -1;

        if (memo[amount] != -6) return memo[amount];

        int res = Integer.MAX_VALUE;
        for (int i = 0; i < coins.length; i++) {
            if (amount < coins[i]) continue;
            int sub = dp(coins, amount - coins[i]);
            if (sub == -1) continue;
            res = Math.min(res, 1 + sub);
        }
        memo[amount] = (res == Integer.MAX_VALUE) ? -1 : res;
        return memo[amount];
    }
}

• 时间复杂度：O(n)
• 空间复杂度：O(n)

解法三：DP table自底向上

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount+1];
        Arrays.fill(dp, amount+1);
        dp[0] = 0;
        int res = Integer.MAX_VALUE;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (i < coin) continue;
                dp[i] = Math.min(dp[i], 1+dp[i-coin]);
            }
        }
        return dp[amount] == amount+1 ? -1 : dp[amount];
    }
}

• 时间复杂度：O(n)
• 空间复杂度：O(n)