673. Number of Longest Increasing Subsequence

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最新推荐文章于 2025-01-13 15:16:10 发布

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本文介绍了一种算法，用于找出未排序整数数组中所有最长递增子序列的数量。通过示例说明了如何处理不同情况，如[1,3,5,4,7]和[2,2,2,2,2]，并详细解释了实现过程。

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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.

class Solution {
public:
   int findNumberOfLIS(vector<int>& nums) {
        int n = nums.size(), max_len = 1, res = 0;
        vector<int> dp(n, 1), cnt(n, 1);   //dp记录以i结尾的最长子序列长度 ，cnt记录以i结尾的最长子序列的个数
        for(int i = 1; i < n; ++i){
            for(int j = 0; j < i; ++j){
                if(nums[j] < nums[i] && dp[j] + 1 > dp[i]){
                    dp[i] = dp[j] + 1;
                    cnt[i] = cnt[j];
                } else if(nums[j] < nums[i] && dp[j] + 1 == dp[i]){
                    cnt[i] += cnt[j];
                }
            }
            max_len = max(max_len, dp[i]);
        }
        for(int i = 0; i < n; ++i)
            if(dp[i] == max_len) res += cnt[i];
        return res;
    }
};