Week 12 - 673.Number of Longest Increasing Subsequence

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;
    }
};
posted @ 2017-11-21 18:31  JerryChan31  阅读(110)  评论(0编辑  收藏  举报