LeetCode-322. Coin Change

LeetCode-322. Coin Change

You are given coins of different denominations and a total amount of money amount. Write a function to compute the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return ​​-1​​.

Example 1:

Input: coins = ​​[1, 2, 5]​​​, amount = ​​11​​ Output: ​​3​​ Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = ​​[2]​​​, amount = ​​3​​ Output: -1

Note: You may assume that you have an infinite number of each kind of coin.

题解：

第一想法就是dp，判断coins组成1-amount的最少硬币数，然后返回dp[n - 1][amount]；

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

发现每次dp[i][j]只用到了dp[i][j - 1]这个值，可以不存储coins这个维度。

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

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