Question

    /* A Naive recursive implementation of LIS problem */
    #include<stdio.h>
    #include<stdlib.h>

    /* To make use of recursive calls, this function must return two things:
       1) Length of LIS ending with element arr[n-1]. We use max_ending_here
          for this purpose
       2) Overall maximum as the LIS may end with an element before arr[n-1]
          max_ref is used this purpose.
    The value of LIS of full array of size n is stored in *max_ref which is our final result
    */
    int _lis( int arr[], int n, int *max_ref)
    {
        /* Base case */
        if(n == 1)
            return 1;

        int res, max_ending_here = 1; // length of LIS ending with arr[n-1]

        /* Recursively get all LIS ending with arr[0], arr[1] ... ar[n-2]. If
           arr[i-1] is smaller than arr[n-1], and max ending with arr[n-1] needs
           to be updated, then update it */
        for(int i = 1; i < n; i++)
        {
            res = _lis(arr, i, max_ref);
            if (arr[i-1] < arr[n-1] && res + 1 > max_ending_here)
                max_ending_here = res + 1;
        }

        // Compare max_ending_here with the overall max. And update the
        // overall max if needed
        if (*max_ref < max_ending_here)
           *max_ref = max_ending_here;

        // Return length of LIS ending with arr[n-1]
        return max_ending_here;
    }

    // The wrapper function for _lis()
    int lis(int arr[], int n)
    {
        // The max variable holds the result
        int max = 1;

        // The function _lis() stores its result in max
        _lis( arr, n, &max );

        // returns max
        return max;
    }

    /* Driver program to test above function */
    int main()
    {
        int arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 };
        int n = sizeof(arr)/sizeof(arr[0]);
        printf("Length of LIS is %d\n",  lis( arr, n ));
        getchar();
        return 0;

令arr [0..n-1]为输入数组，L（i）为LIS的长度，直到索引i，使得arr [i]是LIS的一部分，arr [i]是最后一个元素在LIS中，则L（i）可以递归写为。 L（i）= {1 + Max（L（j））}其中j <1。我和arr [j]＆lt; arr [i]如果没有这样的j则L（i）= 1。

在上述实现中，我无法理解条件if (arr[i-1] < arr[n-1] && res + 1 > max_ending_here)的用途/重要性。它甚至看起来不像递归公式，那么为什么需要它。当L(i)/*is just*/ = { 1 + Max ( L(j) ) } where j < i and arr[j] < arr[i] and if there is no such j then L(i) = 1为什么我们需要比较arr[i-1] < arr[n-1]。是否有可能提供类似于递归公式的递归解决方案？

Answer 1

LIS ：这是遵循LIS定义的简单解决方案。假设 A 是数字输入数组， N 的大小为 A 。

int L[51];
int res=-1;
for(int i=0;i<N;i++)
{
 L[i]=1;
 for(int j=0;j<i;j++)
  if(A[j]<A[i])
  {
    L[i]=max(L[i],L[j]+1);
  }
 res=max(res,L[i]);
}
return res;

时间复杂度：O（N ²）。

最长增长子序列中的荒谬条件

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