Number of Longest Increasing Subsequence

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.

1.解析

题目大意，求解最长子序列的所有个数。

2.分析

整体上看，这道题是最长子序列的升级版。难点就在于如何保存出现多个最长子序列时的个数，参考@Granyang的博客，巧妙的用到了两个dp数组，len[i]表示前i个序列的最长的子序列，cnt[i]表示前i个序列的最长的子序列的个数；如果当前nums[i] <= nums[j]，那么不满足递增子序列的定义，无需考虑；反之，则nums[i]可以作为子序列[0, j]的下一个元素，若当前len[i] == len[j] + 1，说明存在多个当前最长子序列，cnt[i] += cnt[j]；若当前len[i] < len[j] + 1，说明找到目前最长的子序列，cnt[i] = cnt[j]。最后跟全局的最长子序列mx进行比较，并更新状态即可。

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        if (nums.empty()) return 0; //若当前数组为空
        int mx = 1, res = 1, n = nums.size(); //最长子序列的长度至少为1,个数也至少有1个
        vector<int> len(n, 1), cnt(n, 1); 
        for (int i = 1; i < n; ++i){
            for (int j = i - 1; j >= 0; --j){ 
                if (nums[j] < nums[i]){ //满足递增子序列的定义
                   if (len[j] + 1 == len[i]) cnt[i] += cnt[j];
                   else if (len[j] + 1 > len[i]){ //若找到更长的递增子序列，则之前的个数作废，将cnt[j]作为当前最长子序列的个数
                       len[i] = len[j] + 1;
                       cnt[i] = cnt[j];
                   }
                }
            }
            if (mx == len[i]) res += cnt[i]; //跟全局的最长子序列进行比较
            else if (mx < len[i]) mx = len[i], res = cnt[i];
        }
        
        return res;
    }
};

[1]https://www.cnblogs.com/grandyang/p/7603903.html