LeetCode OJ 322. Coin Change DP Solution

LeetCode OJ 322. Coin Change DP Solution
322. Coin ChangeMy SubmissionsQuestionTotal Accepted: 15289 Total Submissions: 62250 Difficulty: Medium

You are given coins of different denominations and a total amount of moneyamount. 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
Return3(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 [email protected] adding this problem and creating all test cases.

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Discuss

A number of coins with fixed nominal values can be used infinitely. The number of targets, which must be exchanged with the minimum number of coins.

Another way of thinking is to understand the question. Each time you can take the number of steps given the nominal value and ask how many steps can be taken to reach the goal. In this way, you can use BFS to solve the problem.

The second solution is DP, dp [I] = min {dp [I-a], dp [I-B], dp [I-c]...}. Dynamic Planning can be quickly solved with formulas.

The AC code I solved using DP

 

Package march; import java. util. arrays;/*** @ auther lvsheng * @ date July 15, March 16, 2016 * @ time 11:20:44 * @ project LeetCodeOJ **/public class CoinChange {public static void main (String [] args) {int [] a = {1, 2, 5}; System. out. println (coinChange (a, 11); int [] B = {2}; System. out. println (coinChange (B, 3);} public static int coinChange (int [] coins, int amount) {int len = coins. length; if (len = 0) return -1; int [] dp = new int [amount + 1]; Arrays. fill (dp, Integer. MAX_VALUE); dp [0] = 0; Arrays. sort (coins); for (int I = 0; I <len & coins [I] <= amount; I ++) dp [coins [I] = 1; for (int I = 1; I <= amount; I ++) {int min = dp [I]; if (min = 1) continue; for (int j = 0; j <len & coins [j] <I; j ++) {int c = coins [j]; int d = dp [I-c]; if (d! = Integer. MAX_VALUE & d <min) min = d + 1;} dp [I] = min;} return dp [amount] = Integer. MAX_VALUE? -1: dp [amount];}