Given an integer array nums, return the number of longest increasing subsequences.
Notice that the sequence has to be strictly increasing.
Example 1:
Input: nums = [1,3,5,4,7] Output: 2 Explanation: The two longest increasing subsequences are [1, 3, 4, 7] and [1, 3, 5, 7].
Example 2:
Input: nums = [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.
Constraints:
1 <= nums.length <= 2000-106 <= nums[i] <= 106
dp, time = O(n ^ 2), space = O(n)
class Solution { public int findNumberOfLIS(int[] nums) { int n = nums.length; if(n <= 1) { return n; } int res = 0, maxLen = 0; int[] lengths = new int[n]; //lengths[i] = length of longest ending in nums[i] int[] counts = new int[n]; //count[i] = number of longest ending in nums[i] for(int i = 0; i < n; i++) { lengths[i] = counts[i] = 1; for(int j = 0; j < i; j++) { if(nums[j] < nums[i]) { if(lengths[j] + 1 == lengths[i]) { counts[i] += counts[j]; } if(lengths[j] + 1 > lengths[i]) { counts[i] = counts[j]; lengths[i] = lengths[j] + 1; } } } if(lengths[i] == maxLen) { res += counts[i]; } if(lengths[i] > maxLen) { maxLen = lengths[i]; res = counts[i]; } } return res; } }
673. Number of Longest Increasing Subsequence - Medium
原文：https://www.cnblogs.com/fatttcat/p/13959449.html