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.

思路：DP，先求最大的递增连续序列的个数，然后再DP到i位置位置形成j个连续子序列的个数

第二次DP要利用第一个DP数组求出来的信息 

package l673;
class Solution {
    public int findNumberOfLIS(int[] nums) {
    	if(nums.length == 0)	return 0;
        int max = 1;
        int[] dp1 = new int[nums.length];
        dp1[0] = 1;
        for(int i=1; i nums[j])
        			dp1[i] = Math.max(dp1[i], 1+dp1[j]);
        	max = Math.max(max, dp1[i]);
        }
        
        int[][] dp = new int[nums.length][1+max];
        for(int i=0; i nums[j])
        				dp[i][k+1] += dp[j][k];
    			}
        	}
        }
        
        int ret = 0;
        for(int i=0; i

这个是O(N^3)的，用Python会TLE，其实后一过程可以优化为O(N^2)，dp2[i]表示以i结尾，LIS长度为dp[i]的个数

class Solution(object):
    def findNumberOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums: return 0
        n = len(nums)
        Max_len = 0
        # max_len[i]:max LIS length end with i, cnt[i]:count of current longest LIS length end with i 
        max_len, cnt = [0]*n, [0]*n
        
        for i in range(n):
            max_len[i] = 1
            for j in range(i):
                if nums[i]>nums[j]:
                    max_len[i] = max(max_len[i], max_len[j]+1)
            Max_len = max(Max_len, max_len[i])
        
        for i in range(n):
            if max_len[i]==1: 
                cnt[i]=1
                continue
            for j in range(i):
                if nums[i]>nums[j]:
                    if max_len[i] == max_len[j]+1:
                        cnt[i] += cnt[j]
                        
        return sum(cnt[i] for i in range(n) if max_len[i]==Max_len)

s=Solution()
print(s.findNumberOfLIS([2,2,2,2,2]))
print(s.findNumberOfLIS([1,3,5,4,7]))
print(s.findNumberOfLIS([1,3,2]))

然后2个循环可以写在一起

class Solution(object):
    def findNumberOfLIS(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        n = len(nums)
        cur_max = ret = 0
        # max_len[i]:max LIS length end with i, cnt[i]:max count of current max LIS length 
        max_len, cnt = [0]*n, [0]*n
        
        for i in range(n):
            max_len[i] = cnt[i] = 1
            for j in range(i):
                if nums[i]>nums[j]:
                    if max_len[i]==max_len[j]+1: cnt[i]+=cnt[j]
                    elif max_len[i]