[LeetCode] 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:

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.

题意：给一个金币数组要求用最少的金币数组成 amount

这种题一看就是DP，关键在于找到转方程

dp[i][j] = min(dp[i-1][j], dp[i][j - coins[i]] + 1) 

做dp题要理解dp中两个维度的含义，做起来就容易一些

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp[] = new int [coins.length + 1][amount + 1];
        // 求最小先初始化最大值，留出dp[x][0]为0就行了
        // 用MAX_VALUE时要注意可能越界否，最好给它减一些数
        for (int i = 0; i < coins.length; i++)
            for (int j = 1; j <= amount; j++)
                dp[i][j] = Integer.MAX_VALUE - 1;
        
        // 先把dp[0][x] 的情况处理的，不然之后转移方程会有越界情况
        for (int j = 1; j <= amount; ++ j) {
            if (j - coins[0] >= 0) {
                dp[0][j] = Math.min(dp[0][j - coins[0]] + 1, dp[0][j]);
            }
        }
        
        // 接下来就容易了，只要有转移方程，在可以判断的情况下就去判断
        for (int i = 1; i < coins.length; i++) {
            for (int j = 1; j <= amount; j++) {
                if (j - coins[i] >= 0) {
                    dp[i][j] = Math.min(dp[i][j - coins[i]] + 1, dp[i - 1][j]);
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        if (dp[coins.length - 1][amount] >= Integer.MAX_VALUE - 1)
            return -1;
        return dp[coins.length - 1][amount];
    }
}