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
leetcode 673. Number of Longest Increasing Subsequence
原文：https://www.cnblogs.com/qinduanyinghua/p/11469124.html