我目前正在研究一种算法,用n个整数序列输出所有排列,这个函数可以正常工作直到6个元素,但是当它超过了我得到的StackOverflowError消息时。我已经阅读了一些关于这个主题的问题,但我发现它是在递归方法具有无限量的调用时发生的,并且在基本情况执行时似乎不是这种情况。
//int[] c parameter receives an array with n elemens to permute
//int i represents the length of the sequence to permute (ai,...,aN)
public boolean isPermutated(int[] c, int i)
{
int max = 0; // the maximum number from a sequence (a1,...,aN) and
int index = 0; // the index where the maximum is
int lessThan; // an index preceded by maximum
//Base case
if(i == 1)
{
return false;
}
// Gives the Maximum element from a sequence (ai,...,aN)
for(int count = c.length - i; count < c.length; count++)
{
if(max < c[count])
{
max = c[count];
index = count;
}
}
// Swap the Maximum from index to index-1
if(max != c[c.length - i])
{
lessThan = c[index-1];
c[index-1] = max;
c[index] = lessThan;
//move each maximum from a sequence (a,...,aN) to the last index ex, (3,2,1)->(2,1,3)
if(i != c.length)
{
while((i-c.length) != 0)
{
for(int count = c.length - (i+1); count < c.length-1; count++)
{
int before;
int after;
before = c[count];
after = c[count+1];
c[count] = after;
c[count+1] = before;
}
i++;
}
}
// print array c elements
for(int e:c)
System.out.print(e);
System.out.println();
isPermutated(c, c.length);
}
else
{
i--;
isPermutated(c, i);
}
return true;
}
我很沮丧,因为我需要10个元素的数组。是否存在具有相同方法签名的伪代码,或者是否有任何方法可以调整堆栈跟踪,因为这应该部分解决问题?
答案 0 :(得分:3)
你的代码不会错过return语句吗?为什么你打电话给isPermutated而不用担心重新计算的价值呢?
您的代码永远不会返回true。