673. Number of Longest Increasing Subsequence

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:

Input: [1,3,5,4,7]
Output: 2
Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: [2,2,2,2,2]
Output: 5
Explanation: The length of longest continuous increasing subsequence is 1, and there are 5 subsequences' length is 1, so output 5.

Note: Length of the given array will be not exceed 2000 and the answer is guaranteed to be fit in 32-bit signed int.

题意：

求最长递增子序列的个数。

思路：

首先对于最长递增子序列的1解法，可以参考https://blog.csdn.net/u013178472/article/details/54926531。至于求个数，可以加一个数组count,count[i]表示到第i个最长递增子序列的个数。可以知道，count[i]初始化为1，对于len[i]==len[j]+1，那么count[i]+=count[j]，因为它的数量得加上前面的，对于len[i]<len[j]+1,那么count[i]=count[j]。最后找到最长的长度，判断len[i]是否等于maxn,等于则总数加上count[i].

代码：

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int []len=new int[nums.length];
        int []count=new int[nums.length];
        for(int i=0;i<nums.length;i++)
        {
            len[i]=1;
            count[i]=1;
        }
        int maxn=1;
        for(int i=1;i<nums.length;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(nums[j]<nums[i])
                {
                    if(len[i]==len[j]+1)
                        count[i]+=count[j];
                    else if(len[i]<len[j]+1)
                    {
                        len[i]=len[j]+1;
                        count[i]=count[j];
                    }
                }
            }
            maxn=Math.max(maxn,len[i]);
        }
        int res=0;
        for(int i=0;i<nums.length;i++)
        {
            if(len[i]==maxn)
                res+=count[i];
        }
        return res;
    }
}