Number of Longest Increasing Subsequence

题目
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.

答案

class Solution {
    public int findNumberOfLIS(int[] nums) {
        if(nums.length == 0) return 0;
        int[] lis = new int[nums.length];
        int[] cnt = new int[nums.length];
        int max_len = 1;
        Arrays.fill(lis, 1);
        Arrays.fill(cnt, 1);

        // Calculate lis and cnt
        for(int i = 1; i < nums.length; i++) {
            int longest_len;
            int count = 0;
            for(int j = 0; j < i; j++) {
                if(nums[j] < nums[i]) {
                    lis[i] = Math.max(lis[i], lis[j] + 1);
                }
            }

            // If lis[i] == 1, cnt[i] stays 1, otherwise we use the loop below to count how many LIS ending at nums[i]
            if(lis[i] > 1) cnt[i]--;
            for(int j = 0; j < i; j++) {
                if(lis[i] == lis[j] + 1 && nums[i] > nums[j]) {
                    cnt[i] += cnt[j];
                }
            }
            max_len = Math.max(max_len, lis[i]);
        }
    
        int num_lis = 0;
        for(int i = 0; i < nums.length; i++) {
            if(lis[i] == max_len)
                num_lis += cnt[i];
        }

        return num_lis;
    }
}