322. Coin Change

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.

思路一：回溯算法，时间复杂度为为指数级，进行剪枝处理，任然TimeOut，程序如下所示：

class Solution {
    private int minLen = Integer.MAX_VALUE;
    public int coinChange(int[] coins, int amount) {
        List<Integer> list = new LinkedList<>();
        backTracing(coins, 0, amount, list);
        return minLen == Integer.MAX_VALUE?-1:minLen;
    }
    
    public void backTracing(int[] coins, int start, int amount, List<Integer> list){
        if (start > coins.length||amount < 0){
            return;
        }
        if (amount == 0){
            minLen = Math.min(minLen, list.size());
            return;
        }
        if (list.size() >= minLen){
            return;
        }
        for (int i = start; i < coins.length; ++ i){
            list.add(coins[i]);
            backTracing(coins, i, amount - coins[i], list);
            list.remove(list.size() - 1);
        }
    }
}

思路2：

DP思路，求解的最长长度为amount，因此设置一个数组dp[amount+1]，对每个长度可行的值进行一个映射，且本题只求长度，不要找到具体的解，程序如下所示：

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