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  • Week 12

    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;
        }
    };
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  • 原文地址:https://www.cnblogs.com/JerryChan31/p/8184131.html
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