Week 12 - 673.Number of Longest Increasing Subsequence

时间:2022-05-26 06:48:40

Week 12 - 673.Number of Longest Increasing Subsequence

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.

my solution:

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]);
}
// find the longest increasing subsequence of the whole sequence
// sum valid counts
for (int i = 0; i < n; i++)
if (len[i] == maxlen) ans += cnt[i];
return ans;
}
};