673. Number of Longest Increasing Subsequence

Description

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.

Solution

DP, time O(n ^ 2), space O(n)

是"300. Longest Increasing Subsequence"的简单扩展。

class Solution {
    public int findNumberOfLIS(int[] nums) {
        int n = nums.length;
        int[] dpLen = new int[n];
        int[] dpCount = new int[n];
        int maxLen = 0;
        int maxCount = 0;
        
        for (int i = 0; i < n; ++i) {
            dpLen[i] = 1;
            dpCount[i] = 1;
            
            for (int j = 0; j < i; ++j) {
                if (nums[i] <= nums[j]) {
                    continue;
                }
                
                if (dpLen[i] < dpLen[j] + 1) {
                    dpLen[i] = dpLen[j] + 1;
                    dpCount[i] = dpCount[j];
                } else if (dpLen[i] == dpLen[j] + 1) {
                    dpCount[i] += dpCount[j];   
                }
            }
            
            if (dpLen[i] > maxLen) {
                maxLen = dpLen[i];
                maxCount = dpCount[i];
            } else if (dpLen[i] == maxLen) {
                maxCount += dpCount[i];
            }
        }
        
        return maxCount;
    }
}