LeetCode Number of Longest Increasing Subsequence

原题链接在这里：https://leetcode.com/problems/number-of-longest-increasing-subsequence/description/

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.

题解：

DP. longestLen[i]记录 到i的LIS长度. longestCount[i]记录到i的LIS个数.

Base Case 都是1.

更新, 对每个j<i, 若nums[i] > nums[j], 正常longestLen[j]+1应该是新的longestLen[i]的长度, 但若是已经相等说明之前已经有线能加到这个长度了，所以longestCount[i]就要加上longestCount[j].

result. 每个i维护最长LIS长度，若更新, 把longestCount[i]赋值给res. 若遇到相同的LIS长度，把longestCount[i]加到res中.

Time Complexity: O(n^2), n = nums.length.

Space: O(n).

AC Java:

 1 class Solution {
 2     public int findNumberOfLIS(int[] nums) {
 3         if(nums == null || nums.length == 0){
 4             return 0;
 5         }
 6
 7         int len = nums.length;
 8         int [] longestLen = new int[len];
 9         int [] longestCount = new int[len];
10         int res = 0;
11         int maxLen = 0;
12         for(int i = 0; i<nums.length; i++){
13             longestLen[i] = 1;
14             longestCount[i] = 1;
15             for(int j = 0; j<i; j++){
16                 if(nums[j] < nums[i]){
17                     if(longestLen[j]+1 > longestLen[i]){
18                         longestLen[i] = longestLen[j]+1;
19                         longestCount[i] = longestCount[j];
20                     }else if(longestLen[j]+1 == longestLen[i]){
21                         longestCount[i] += longestCount[j];
22                     }
23                 }
24             }
25
26             if(longestLen[i] > maxLen){
27                 maxLen = longestLen[i];
28                 res = longestCount[i];
29             }else if(longestLen[i] == maxLen){
30                 res += longestCount[i];
31             }
32         }
33         return res;
34     }
35 }

是Longest Continuous Increasing Subsequence 的进阶题.