思路:
动态规划解题,记当前最长递增子序列的最后一个数为a[i],长度为c[i];
记a[j]为a[0]到a[i]中所有以j结尾的递增子序列,如果a[i+1].>a[i],,则c[i+1] = MAX(c[i+1],c[j]+1);
#include<stdlib.h>
int a[8] = {1,2,-2,3,2,3,4,5};
int c[100] = {1};
int max(int a, int b)
{
return a > b ? a : b;
}
int findMaxSub(int* a, int n)
{
int i, j;
for(i = 0; i < n; i++)
{
for(j = 0; j < i; j++)
{
if(a[i] > a[j])
{
c[i] = max(c[i],c[j] + 1);
}
}
}
printf("%d",c[n-1]);
return c[n-1];
}
int main()
{
findMaxSub(a,8);
}