leetcode 673. Number of Longest Increasing Subsequence 动态规划DP

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.

本题基本上和之前的是一样的做法

dp[i]定义为以nums[i]为结尾的递推序列的个数的话，再配上这些递推序列的长度，将会比较容易的发现递推关系。这里我们用len[i]表示以nums[i]为结尾的递推序列的长度，用cnt[i]表示以nums[i]为结尾的递推序列的个数，初始化都赋值为1，只要有数字，那么至少都是1。然后我们遍历数组，对于每个遍历到的数字nums[i]，我们再遍历其之前的所有数字nums[j]，当nums[i]小于等于nums[j]时，不做任何处理，因为不是递增序列。反之，则判断len[i]和len[j]的关系，如果len[i]等于len[j] + 1，说明nums[i]这个数字可以加在以nums[j]结尾的递增序列后面，并且以nums[j]结尾的递增序列个数可以直接加到以nums[i]结尾的递增序列个数上。如果len[i]小于len[j] + 1，说明我们找到了一条长度更长的递增序列，那么我们此时奖len[i]更新为len[j]+1，并且原本的递增序列都不能用了，直接用cnt[j]来代替。我们在更新完len[i]和cnt[i]之后，要更新mx和res，如果mx等于len[i]，则把cnt[i]加到res之上；如果mx小于len[i]，则更新mx为len[i]，更新结果res为cnt[i]

建议和leetcode 300. Longest Increasing Subsequence 最长递增子序列LISS 一起学习

代码如下：

#include <iostream>#include <vector>#include <map>#include <set>#include <queue>#include <stack>#include <string>#include <climits>#include <algorithm>#include <sstream>#include <functional>#include <bitset>#include <numeric>#include <cmath>#include <regex>using namespace std;class Solution{public:    int findNumberOfLIS(vector<int>& nums)     {        vector<int> dp(nums.size(), 1);        vector<int> count(nums.size(), 1);        int maxLen = 0;        for (int i = 0; i < nums.size(); i++)        {            for (int j = 0; j < i; j++)            {                if (nums[j] < nums[i])                {                    if (dp[i] == dp[j] + 1)                        count[i] += count[j];                    else if (dp[i] < dp[j] + 1)                    {                        dp[i] = dp[j] + 1;                        count[i] = count[j];                    }                }               }            maxLen = max(maxLen, dp[i]);        }        int res = 0;        for (int i = 0; i < nums.size(); i++)        {            if (maxLen == dp[i])                res += count[i];        }        return res;    }};