[LeetCode] 322. Coin Change change exchange

topic:

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 In just saw the title, it is natural to think of using dfs, exhaust all possibilities, although it is possible to know a timeout, but still ... really do overtime

Then read the solution to a problem, dynamic programming (bottom-up) particularly clear

analysis:

Definition of F (i) represents a minimum number of required composition i dollars

So there,

F(0)=0

The ultimate goal is to obtain F (amount)

Transition between different states is a formula 　  F. (I) = min (F. (Cj-i)) + . 1 , 0 = <j <coins.size ()  　a j-th where Cj representative of the value of the denomination of the coin, if there i -Cj <0, simply ignored

When seeking F. (I), only you need to know each F (i-Cj), and F (i-Cj) until i is known, and it can be calculated

Code:

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> F(amount+1, amount);
        F[0] = 0;
        for(int i=1; i<=amount; i++){
            int cur=INT_MAX;
            for(int j=0; j<coins.size(); j++){
                if(i-coins[j]<0)    continue;
                else if(F[i-coins[j]]>=0 && cur > F[i-coins[j]] )   cur = F[i-coins[j]]+1;
            }
            F[i] = cur == INT_MAX?-1:cur;
        }
        return F[amount];
    }
};