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  • 换钱最少货币数

    给定数组arr,arr中所有的值都为正数且不重复。每个值代表一种面值的货币,每种面值的货币可以使用任意张,在给定一个整数aim代表要找的钱数,求组成aim的最少货币数。

    代码:

    public class MinCoins {
        public static int minCoins(int[] arr, int aim) {
            if(arr == null || arr.length==0 || aim<=0) {
                return 0;
            }
            
            int len = arr.length;
            int max = Integer.MAX_VALUE;
            int[][] dp = new int[len][aim+1];
            //初始化数组
            for(int j=1; j<(aim+1); j++) {
                dp[0][j] = max;
                if((j-arr[0])>=0 && dp[0][j-arr[0]]!=max) {
                    dp[0][j]= dp[0][j-arr[0]]+1;
                }
            }
            
            int left=0;
            for(int i=1; i<len; i++) {
                for(int j=1; j<(aim+1); j++) {
                    left = max;
            //重点在dp[i][j-arr[i]],不是dp[i-1][j-arr[i]],这是算法推到出来的结果
                    if((j-arr[i])>=0 && dp[i][j-arr[i]]!=max) {
                        left = dp[i][j-arr[i]]+1;
                    }
                    dp[i][j] = Math.min(dp[i-1][j], left);
                }
            }
            
            for (int[] rows : dp) {
              for (int col : rows) {
                  System.out.format("%5d", col);
              }
              System.out.println();
         }
            
            return dp[len-1][aim] == max ? -1 :dp[len-1][aim];
        }
        
        public static void main(String[] args) {
            int[] arr = {5,2,3};
            System.out.println(minCoins(arr,20));
        }
    }

    给定数组arr,arr中所有的值都为正数且不重复。每个值代表一种面值的货币,在给定一个整数aim代表要找的钱数,求组成aim的最少货币数。

    注意:此处只能用数组中的数,并且每个只能用一次。

    代码:

    public class MinCoins {
        public static int minCoins(int[] arr, int aim) {
            if(arr == null || arr.length==0 || aim<=0) {
                return -1;
            }
            
            int len = arr.length;
            int max = Integer.MAX_VALUE;
            int[][] dp = new int[len][aim+1];
            
            //初始化数组
            for(int j=1; j<(aim+1); j++) {
                dp[0][j] = max;
                if((j-arr[0])==0) {
                    dp[0][j]= 1;
                }
            }
            
            int left=0;
            for(int i=1; i<len; i++) {
                for(int j=1; j<(aim+1); j++) {
                    left = max;
                    if((j-arr[i])>=0 && dp[i-1][j-arr[i]]!=max) {
                        left = dp[i-1][j-arr[i]]+1;
                    }
                    dp[i][j] = Math.min(dp[i-1][j], left);
                }
            }
            
            for (int[] rows : dp) {
              for (int col : rows) {
                  System.out.format("%5d", col);
              }
              System.out.println();
         }
            
            return dp[len-1][aim]==max ? -1 : dp[len-1][aim];
        }
        
        public static void main(String[] args) {
            int[] arr = {5,2,3,5};
            System.out.println(minCoins(arr,15));
        }
    }
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  • 原文地址:https://www.cnblogs.com/loren-Yang/p/7482508.html
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