673. Number of Longest Increasing Subsequence

Posted 2020-10-20 小预备

#week11

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.

 

分析

这个和平常的最长子序列的不同的点在于是要求该子序列的个数

因此需要在DP时再加一维用来存放个数

初始化每个状态的每个维都为1

状态变换：

if(dp[i].first == dp[j].first + 1)dp[i].second += dp[j].second;
if(dp[i].first < dp[j].first + 1)dp[i] = {dp[j].first + 1, dp[j].second};

 

题解

 1 class Solution {
 2 public:
 3     int findNumberOfLIS(vector<int>& nums) {
 4         int n = nums.size(), res = 0, max_len = 0;
 5         vector<pair<int,int>> dp(n,{1,1});            //dp[i]: {length, number of LIS which ends with nums[i]}
 6         for(int i = 0; i<n; i++){
 7             for(int j = 0; j <i ; j++){
 8                 if(nums[i] > nums[j]){
 9                     if(dp[i].first == dp[j].first + 1)dp[i].second += dp[j].second;
10                     if(dp[i].first < dp[j].first + 1)dp[i] = {dp[j].first + 1, dp[j].second};
11                 }
12             }
13             if(max_len == dp[i].first)res += dp[i].second;
14             if(max_len < dp[i].first){
15                 max_len = dp[i].first;
16                 res = dp[i].second;
17             }
18         }
19         return res;
20     }
21 };