问题描述：
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.
解题思路：
可以用一个数组lens存储到这个数字位置最长的自增序列的长度，用一个数组cnt存储以这个数字为结尾的最长的递增序列的个数。
更新lens比较好更新：比较nums[i] 和 nums[j]的大小。
更新cnt时需注意，需要比较lens[i] 和 lens[j]的大小：
注意此时前提是nums[j] < nums[i]
1. 若lens[i] < lens[j] + 1 也就是说 lens[i] 记录的不为最长的，至少比lens[j] , nums[i]的序列要短
此时我们更新lens[i] = lens[j] +1, cnt[i] = cnt[j]
2.若lens[i] == lens[j] + 1，说明当前这个序列又是满足条件的最长子序列 所以 cnt[i] += cnt[j]
代码：
class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
vector<int> lens(nums.size(), 1);
vector<int> cnt(nums.size(), 1);
int maxLen = 0;
for(int i = 0; i < nums.size(); i++){
for(int j = 0; j < i; j++){
if(nums[i] > nums[j]){
if(lens[i] == lens[j] + 1)
cnt[i] += cnt[j];
else if(lens[i] < lens[j] + 1){
lens[i] = lens[j]+1;
cnt[i] = cnt[j];
}
}
}
maxLen = max(maxLen, lens[i]);
}
int ret = 0;
for(int i = 0; i <lens.size(); i++){
if(lens[i] == maxLen)
ret += cnt[i];
}
return ret;
}
};
来源：https://www.cnblogs.com/yaoyudadudu/p/9197364.html