完全背包

518. 零钱兑换 II 

遍历硬币和金额，累加所有可能

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

此处涉及到排列情况，需要先遍历大小，再遍历容量

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