代码随想录算法训练营第三七天| 动态规划：完全背包理论基础 518.零钱兑换II 377. 组合总和 Ⅳ 322. 零钱兑换

今日任务

动态规划：完全背包理论基础
518.零钱兑换II
377. 组合总和 Ⅳ
322. 零钱兑换

518.零钱兑换II

题目链接： . - 力扣（LeetCode）

class Solution {public int change(int amount, int[] coins) {int[] dp = new int[amount + 1];dp[0] = 1;for (int i = 0; i < coins.length; i++) {for (int j = coins[i]; j <= amount; j++) {dp[j] += dp[j - coins[i]];}}return dp[amount];}
}

377. 组合总和 Ⅳ

题目链接： . - 力扣（LeetCode）

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

322. 零钱兑换

题目链接： . - 力扣（LeetCode）

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