假设我们有一个数字列表。我们必须找到最长的双子序列的长度。正如我们所知,如果严格地增加然后严格地减少,则该序列被称为二元序列。严格增加的顺序是双声的。或严格减少的顺序也是双音的。
因此,如果输入类似于nums = [0、8、4、12、2、10、6、14、1、9、5、13、11、7、15],序列16的大小,则输出将是7。
为了解决这个问题,我们将遵循以下步骤-
增加SubSeq:=给定数组大小的新数组,并用1填充
对于初始化i:= 1,当i <size时,更新(将i增加1),执行-
对于初始化j:= size-1,当j> i时,更新(将j减1),-
对于初始化i:= 1,当i <size时,更新(将i增加1),执行-
reductionSubSeq [i]:= reductionSubSeq [j] + 1
如果arr [i]> arr [j],而reductionSubSeq [i] <reductionSubSeq [j] + 1,则-
最大:=递增SubSeq [0] +递减SubSeq [0]-1
最大值:=递增SubSeq [i] +递减SubSeq [i]-1
如果递增SubSeq [i] +递减SubSeq [i]-1>最大值,则:
最大回报
如果arr [i]> arr [j]且递增的SubSeq [i] <递增的SubSeq [j] + 1,则-
* decreasingSubSeq:=给定数组大小的新数组,并用1填充
递增SubSeq [i]:=增加SubSeq [j] + 1
对于初始化j:= 0,当j <i时,更新(将j增加1),执行-
对于初始化i:= size-2,当i> = 0时,更新(将i减1),执行-
让我们看下面的实现以更好地理解-
#include<iostream> using namespace std; int longBitonicSub( int arr[], int size ) { int *increasingSubSeq = new int[size]; for (int i = 0; i < size; i++) increasingSubSeq[i] = 1; for (int i = 1; i < size; i++) for (int j = 0; j < i; j++) if (arr[i] > arr[j] && increasingSubSeq[i] < increasingSubSeq[j] + 1) increasingSubSeq[i] = increasingSubSeq[j] + 1; int *decreasingSubSeq = new int [size]; for (int i = 0; i < size; i++) decreasingSubSeq[i] = 1; for (int i = size-2; i >= 0; i--) for (int j = size-1; j > i; j--) if (arr[i] > arr[j] && decreasingSubSeq[i] < decreasingSubSeq[j] + 1) decreasingSubSeq[i] = decreasingSubSeq[j] + 1; int max = increasingSubSeq[0] + decreasingSubSeq[0] - 1; for (int i = 1; i < size; i++) if (increasingSubSeq[i] + decreasingSubSeq[i] - 1 > max) max = increasingSubSeq[i] + decreasingSubSeq[i] - 1; return max; } int main() { int arr[] = {0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15}; int n = 16; cout << longBitonicSub(arr, n); }
[0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15], 16
输出结果
7