leetcode 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.

题目大意：求最长递增子序列的数量。

方法一：动态规划

 1 class Solution {
 2 public:
 3     int findNumberOfLIS(vector<int>& nums) {
 4         int len = nums.size();
 5         vector<int> dp(len, 1); // dp[i] = length of longest ending in nums[i]
 6         vector<int> count(len, 1); // count[i] = number of longest ending in nums[i]
 7         int maxn = 0;
 8         for (int i = 0; i < len; i++) {
 9             for (int j = 0; j < i; j++) {
10                 if (nums[i] > nums[j]) {
11                     if (dp[j] >= dp[i]) {
12                         dp[i] = dp[j] + 1;
13                         count[i] = count[j];
14                     } else {
15                         if (dp[j] + 1 == dp[i])
16                             count[i] += count[j];
17                     }
18                 }
19             }
20             maxn = max(maxn, dp[i]);
21         }
22         int cnt = 0;
23         for (int i = 0; i < len; i++) {
24             if (dp[i] == maxn)
25                 cnt += count[i];
26         }
27         return cnt;
28     }
29 };

方法二：线段树

https://leetcode.com/articles/a-recursive-approach-to-segment-trees-range-sum-queries-lazy-propagation/