T. *_*mas 6 java dynamic-programming greedy
我无法弄清楚动态硬币更改问题的最后一段代码.我已经包含了以下代码.
我无法弄清楚最后一个else.我应该在那时使用贪婪算法,还是可以从表中已有的值计算答案?我努力尝试理解这个问题,我觉得我非常接近.该方法通过创建表并使用存储在表中的结果来解决较大的问题而不使用递归,从而找到进行特定更改所需的最小硬币数量.
public static int minCoins(int[] denom, int targetAmount){
int denomPosition; // Position in denom[] where the first spot
// is the largest coin and includes every coin
// smaller.
int currentAmount; // The Amount of money that needs to be made
// remainingAmount <= initalAmount
int[][] table = new int[denom.length][targetAmount+1];
for(denomPosition = denom.length-1 ; denomPosition >= 0 ; denomPosition--) {
for(currentAmount = 0 ; currentAmount <= targetAmount ; currentAmount++){
if (denomPosition == denom.length-1){
table[denomPosition][currentAmount] =
currentAmount/denom[denomPosition];
}
else if (currentAmount<denom[denomPosition]){
table[denomPosition][currentAmount] =
table[denomPosition+1][currentAmount];
}
else{
table[denomPosition][currentAmount] =
table[denomPosition+1][currentAmount]-
table[denomPosition][denom[denomPosition]]-1;
}
}
}
return table[0][targetAmount];
}
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解决换硬币问题不需要改用贪心算法,可以用动态规划算法来解决。例如,像这样:
public int minChange(int[] denom, int targetAmount) {
int actualAmount;
int m = denom.length+1;
int n = targetAmount + 1;
int inf = Integer.MAX_VALUE-1;
int[][] table = new int[m][n];
for (int j = 1; j < n; j++)
table[0][j] = inf;
for (int denomPosition = 1; denomPosition < m; denomPosition++) {
for (int currentAmount = 1; currentAmount < n; currentAmount++) {
if (currentAmount - denom[denomPosition-1] >= 0)
actualAmount = table[denomPosition][currentAmount - denom[denomPosition-1]];
else
actualAmount = inf;
table[denomPosition][currentAmount] = Math.min(table[denomPosition-1][currentAmount], 1 + actualAmount);
}
}
return table[m-1][n-1];
}
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