322. Coin Change

最新推荐文章于 2024-03-19 21:22:22 发布

且听疯吟

最新推荐文章于 2024-03-19 21:22:22 发布

阅读量186

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/mml1106/article/details/77953615

leetcode 专栏收录该内容

84 篇文章

订阅专栏

本文介绍了一种计算组成特定金额所需的最少硬币数量的算法。通过动态规划方法，该算法能在给定不同面额硬币的情况下，找出最少的硬币组合数。若无法组成指定金额，则返回-1。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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:
coins = [1, 2, 5], amount = 11
return 3 (11 = 5 + 5 + 1)

Example 2:
coins = [2], amount = 3
return -1.

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

Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

/*c++ solution*/
class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int Max = amount + 1;
        vector<int> dp(amount+1,Max);
        dp[0] = 0;

        for(int i = 1; i <= amount; ++i ){
            for(int j = 0; j < coins.size(); ++j ){
                if(i >= coins[j]){
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }

        return dp[amount] > amount?-1:dp[amount];
    }
};

/*本人自己写的代码，C++为论坛给出的答案*/
int coinChange(int* coins, int coinsSize, int amount) {
    int *min = NULL;
    int minCoin = 0;
    int i,j;

    if(coins == NULL || coinsSize <= 0 || amount < 0){
        return -1;
    }

    if(amount == 0){
        return 0;
    }

    /*initial the element of the array*/
    min = (int *)malloc(sizeof(int)*(amount+1));
    min[0] = 0;
    minCoin = coins[0];
    for(j = 0;j < coinsSize;++j){
        if(coins[j] < minCoin){
            minCoin = coins[j];
        }
    }  
    if(minCoin > amount){
        return -1;
    }
    for(j = 1;j < minCoin;j++){
        min[j] = -1;
    }

    /*get the min coins*/
    for( i = minCoin;i <= amount; ++i){
        min[i] = -1;
        for(j = 0;j < coinsSize;++j){
            if(i == coins[j]){
                min[i] = 1;
            }else{
                if(i > coins[j] && min[i- coins[j]] > 0 ){
                    if(min[i] == -1){
                        min[i] = (min[i- coins[j]] + 1);
                    }else{
                        min[i] = min[i] < (min[i- coins[j]] + 1)?min[i]:(min[i- coins[j]] + 1);
                    }                
                }
            }           
        }
    }

    return min[amount];
}