一维数组和二维数组求解硬币组合个数问题

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1.利用二维数组求解硬币组合问题

private static long countWays_2(int n) {int[] coins = {1,5,10,25};//二维数组的方式long[][] dp = new long[7][n+1];for (int i = 1; i <=n; i++) {dp[0][i] = 0;//用0种硬币凑成i元的组合数为0}for (int i = 0; i < coins.length; i++) {dp[i][0] = 1;//用i种硬币凑成0元的组合数为1}for (int i = 1; i <= coins.length; i++) {for (int j = 1; j <= n; j++) {dp[i][j]= 0;for (int k = 0; k <=j/coins[i-1]; k++) {dp[i][j] += dp[i-1][j-k*coins[i-1]];//dp[i][sum] = dp[i-1][sum - 0*Vm] + ... + dp[i-1][sum - K*Vm]; 其中K = sum / Vm}}}return dp[coins.length][n];}

2.利用一维数组求解硬币组合问题

private static int countWays(int n) {int[] coins = {1,5,10,25};//一维数组int[] dp = new int[100001];//默认为0dp[0] = 1;int len1 =  coins.length;for (int i = 0; i < len1; i++) {for (int j = coins[i]; j <= n; j++) {//二级循环从大到小dp[j] = (dp[j]+dp[j-coins[i]])%100000007;}}for (int i = 0; i < dp.length; i++) {if(dp[i]!=0)System.out.print(dp[i]+" ");}System.out.println();return dp[n];}

注：何时使用一维数组和二维数组求解动态规划问题可以参考：http://blog.csdn.net/tbwood/article/details/22747215

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