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;

    }

};

巴特西

Week 12 - 673.Number of Longest Increasing Subsequence

Week 12 - 673.Number of Longest Increasing Subsequence

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