322.零钱兑换

思路

• dp问题，空间换时间，递推公式(初始化+转移方程),

F(S) = F(S-C) + 1

# S 代表总额(amount), F(S)代表最少兑换次数，C代表兑换的最后一个面值，其中 S为0时，F(S) = 0, 零钱数组为空时，F(S)=-1.

解法

• 暴力穷举，回溯

public int coinChange(int[] coins, int amount) {
        return coinChange(0, coins, amount);
    }

    private int coinChange(int coinsIndex, int[] coins, int amount) {
        if (amount == 0) {
            return 0;
        }
        if (coinsIndex == coins.length) {
            return -1;
        }
        int min = Integer.MAX_VALUE;
        int maxNum = amount / coins[coinsIndex];
        for (int num = 0; num <= maxNum; num++) {
            int otherNums = coinChange(coinsIndex + 1, coins, amount - num * coins[coinsIndex]);
            if(otherNums != -1) {
                min = Math.min(min, otherNums + num);
            }
        }
        return min == Integer.MAX_VALUE ? -1 : min;
    }

• dp, 自上而下,递归

   public int coinChange(int[] coins, int amount) {
        return coinChange(coins, amount, new int[amount]);
    }

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

• dp, 自下而上，迭代

    public int coinChange(int[] coins, int amount) { 
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int lastCoin : coins) {
                if (i >= lastCoin) {
                    dp[i] = Math.min(dp[i], dp[i - lastCoin] + 1);
                }
            }
        }
        if (dp[amount] == amount + 1) {
            return -1;
        }
        return dp[amount];
    }

• 参考: https://leetcode.com/articles/coin-change/

来源：https://www.cnblogs.com/lasclocker/p/11335783.html