322. Coin Change

You is given coins of different denominations and a total amount of money Amount. Write a function to compute the fewest number of coins so need to make up that amount. If that amount of cannot is 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 so you have a infinite number of each kind of coin.

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

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Dp:constructing the solution in the order of sum. (Get 1, 2, ..., amount)

classSolution { Public intCoinchange (int[] coins,intamount) {    if(Amount < 1)      return0; int[] DP =New int[Amount + 1]; intsum = 0;  while(++sum <=amount) {      //constructing the result from ..., amount      intmin =-1;//-1 means cannot get this amount.       for(intcoin:coins) {        if(Sum >= coin && dp[sum-coin]! =-1) {          intTempsolution = Dp[sum-coin] + 1; if(min = =-1) min= Tempsolution;//Initialize solution.          Elsemin=math.min (tempsolution, Min); }} Dp[sum]=min; }    returnDp[amount]; }}

Dp:constructing the solution by how many coins used. (use 1 coin, use 2 coins, ... until the target is reached.)

 Public classSolution { Public intCoinchange (int[] coins,intamount) {        if(Amount = = 0)            return0; intLen = amount+1; int[] result =New int[Len]; result[0] = 0; List<Integer> q =NewArraylist<integer>(); intMinimumcoins = 1;  for(Integer c:coins) {if(C <Len) {Result[c]=Minimumcoins;            Q.add (c); }        }                 while(!Q.isempty ()) {            ++Minimumcoins; if(Result[amount]! = 0)                returnResult[amount]; List<Integer> next =NewArraylist<integer>();  for(Integer c:coins) { for(Integer optimal:q) {intsum = c+Optimal; if(Sum < len && result[sum] = = 0)                    {                        //Find The new numbers so you can get a with minimumcoins coins. //Add One more coin to these numbers,//You can get the next group of those optimial numbers.Result[sum] =Minimumcoins;                    Next.add (sum); }}} Q=Next; }                return-1; }}

322. Coin Change