Leetcode | Number of Longest Increasing Subsequence

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

原题意内容如下：

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:
Input: [1,3,5,4,7]
Output: 2

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.

题目描述的大概意思就是给定一个无序数组nums，然后找出这个数组最长递增子序列的个数。题目刚开始做的时候，我就隐约感觉这是要用动态规划方程做的，但一直没分析出子问题和原问题之间的关系，即nums[i]和nums[i+1]的对应的最大子序列长度和其个数的关系。

后面开始用len[i]代表到nums[i]结束的最长子序列的长度，用cnt[i]代表最长子序列的个数，然后存在以下两个方程：

len[k+1] = max(len[k+1], len[i]+1) for all i <= k and nums[i] < nums[k+1];

cnt[k+1] = sum(cnt[i]) for all i <= k and nums[i] < nums[k+1] and len[i] = len[k+1]-1;

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根据上述两个动态规划方程，即可进行编程，以下为源代码：

class Solution {public:    int findNumberOfLIS(vector<int>& nums) {        int n = nums.size(), maxlen = 1, ans = 0;        vector<int> cnt(n, 1), len(n, 1);        for (int i = 1; i < n; i++) {            for (int j = 0; j < i; j++) {                if (nums[i] > nums[j]) {                    if (len[j]+1 > len[i]) {                        len[i] = len[j]+1;                        cnt[i] = cnt[j];                    }                    else if (len[j]+1 == len[i])                         cnt[i] += cnt[j];                }            }            maxlen = max(maxlen, len[i]);        }        for (int i = 0; i < n; i++)             if (len[i] == maxlen) ans += cnt[i];        return ans;    }};