@@ -756,6 +756,66 @@ func min(a,b int)int{
756756
757757> dp\[ i-a\[ j]] 决策 a\[ j] 是否参与
758758
759+ Coin Change 2
760+ 761+ ``` cpp
762+ // Some code
763+ class Solution {
764+ public:
765+ int change(int amount, vector<int >& coins) {
766+ 767+ // Time O(n* m) Space O(n* m) Cost: 15min.
768+ /// dp[ i] [ j ] chose from 0 to i coins and get target amount j. combination
769+ // dp[ i] [ j ] dp[ i] [ j ] = dp[ i-1] [ j ] + loop k from 0 to max num coins[ i] . dp[ i-1] [ j-coins[ i] * k]
770+ // dp[ i] [ j ] dp[ i] [ j ] = dp[ i-1] [ j ] + dp[ i] [ j - coins[ i]] ;
771+ // we can improve it further to optimize memory to O(n)
772+ 773+ vector<int > dp(amount+1, 0);
774+ dp[ 0] =1;
775+ for(int i=0; i<coins.size(); i++)
776+ {
777+ 778+ for(int j=coins[ i] ; j <=amount; j++)
779+ {
780+ dp[ j] +=dp[ j-coins[ i]] ;
781+ }
782+ }
783+ return dp[ amount] ;
784+ }
785+ 786+ int method1(int amount, vector<int >& coins)
787+ {
788+ vector<vector<int >> dp(coins.size()+1, vector<int >(amount+1, 0));
789+ 790+ for(int i=0; i<=coins.size(); i++)
791+ {
792+ for(int j=0; j<=amount; j++)
793+ {
794+ if(j==0)
795+ dp[ i] [ j ] = 1;
796+ else if(i==0)
797+ dp[ i] [ j ] = 0;
798+ else
799+ {
800+ int k=0;
801+ /* while(j >=coins[ i-1] * k)
802+ {
803+ dp[ i] [ j ] +=dp[ i-1] [ j-coins[ i-1] * k] ;
804+ k++;
805+ } * /
806+ if(j-coins[ i-1] >=0)
807+ dp[ i] [ j ] = dp[ i-1] [ j ] + dp[ i] [ j-coins[ i-1]] ;
808+ else
809+ dp[ i] [ j ] = dp[ i-1] [ j ] ;
810+ }
811+ }
812+ }
813+ 814+ return dp[ coins.size()] [ amount ] ;
815+ }
816+ };
817+ ```
818+
759819### [backpack](https://www.lintcode.com/problem/backpack/description)
760820
761821> 在 n 个物品中挑选若干物品装入背包,最多能装多满?假设背包的大小为 m,每个物品的大小为 A\[i]
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