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];}
};