最長遞增子序列

標籤：最長遞增子序列   c   二分尋找   

1. 動態規劃，使用一個數組儲存當前的最大遞增子序列長度，時間複雜度為O(N^2)

# include <iostream># include <cstdlib># include <climits>using namespace std;int longestsub(int a[],int n){int *dis=(int *)malloc((n+1)*sizeof(int));dis[0]=1;int i,j;for(i=0;i<n;i++)dis[i]=1;for(i=0;i<n;i++){for(j=i-1;j>=0;j--){if(a[i]>a[j]&&dis[i]<dis[j]+1){dis[i]=dis[j]+1;}}}int ans=-1;for(i=0;i<n;i++)ans=max(ans,dis[i]);free(dis);return ans;}int main(){int a[5]={3,6,5,4,7};cout<<longestsub2(a,5)<<endl;    system("pause");    return 0;}

2.《編程之美》上提供了另外一種解法，使用數組b[len-1]表示長度為len時最後一個元素值，在這種解法中可以使用二分尋找使得程式加速，時間複雜度變為O(N*logN)

# include <iostream># include <cstdlib># include <climits>using namespace std;int binsearch(int a[],int n,int target)  //binarysearch{int low=0;int high=n-1;int mid;while(low<=high){mid=(high-low)/2+low;if(a[mid]==target)return mid;else if(a[mid]<target)low=mid+1;else high=mid-1;}return low;}int longestsub2(int a[],int n){int i=0;int len=1;int *b=(int *)malloc((n+1)*sizeof(int));b[0]=a[0];for(i=1;i<n;i++){if(a[i]>b[len-1]){b[len]=a[i];len++;}else {int tmp=binsearch(a,n,a[i]);b[tmp]=a[i];}}free(b);return len;}int main(){int a[5]={7,6,8,4,1};cout<<longestsub2(a,5)<<endl;    system("pause");    return 0;}