518.零钱兑换 II

要点：先遍历数组，后遍历背包，求的是组合数

class Solution {
public:
    int change(int amount, vector<int>& coins) {
        vector<int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int i = 0; i < coins.size(); i++) {
            for (int j = coins[i]; j <= amount; j++) {
                dp[j] += dp[j - coins[i]];
            }
        }
        return dp[amount];
    }
};

377.组合总和 Ⅳ

要点：先遍历背包，后遍历数组，求的是排列数

class Solution {
public:
    int combinationSum4(vector<int>& nums, int target) {
        vector<int> dp(target + 1, 0);
        dp[0] = 1;
        for (int j = 0; j <= target; j++) {
            for (int i = 0; i < nums.size(); i++) {
                if (j >= nums[i] && dp[j] < INT_MAX - dp[j - nums[i]]) {
                    dp[j] += dp[j - nums[i]];
                }
            }
        }
        return dp[target];
    }
};