365天挑战LeetCode1000题——Day 022 动态规划 20（入门完结）

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322. 零钱兑换（120）

代码实现（首刷自解）

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        vector<int> dp(amount + 1, 10001);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < n; j++) {
                if (i - coins[j] >= 0 && dp[i - coins[j]] != 10001)
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
            }
        }
        return dp[amount] == 10001 ? -1 : dp[amount];
    }
};

518. 零钱兑换 II（121）

代码实现（看题解）

考虑硬币组合的顺序，达到去重的效果

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

377. 组合总和 Ⅳ（122）

代码实现（首刷自解）

class Solution {
public:
    int combinationSum4(vector<int>& coins, int amount) {
        int n = coins.size();
        vector<unsigned int> dp(amount + 1, 0);
        dp[0] = 1;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (i - coin >= 0) {
                    dp[i] += dp[i - coin];
                }
            }
        }
        return dp[amount];
    }
};

343. 整数拆分（123）

代码实现（首刷自解）

class Solution {
public:
    int integerBreak(int n) {
        if (n <= 3) return n - 1;
        vector<int> dp(n + 1, 0);
        dp[1] = 1;
        dp[2] = 2;
        dp[3] = 3;
        for (int i = 4; i <= n; i++) {
            for (int j = 1; j <= i / 2; j++) {
                dp[i] = max(dp[i - j] * dp[j], dp[i]);
            }
        }
        return dp[n];
    }
};

279. 完全平方数（124）

代码实现（首刷自解）

class Solution {
public:
    int numSquares(int n) {
        vector<int> coins;
        for (int i = 1; i <= sqrt(n); i++) {
            coins.push_back(i * i);
        } 
        int amount = n;
        int m = coins.size();
        vector<int> dp(amount + 1, 10001);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < m; j++) {
                if (i - coins[j] >= 0 && dp[i - coins[j]] != 10001)
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
            }
        }
        return dp[amount] == 10001 ? -1 : dp[amount];
    }
};