LeeCode673. Number of Longest Increasing Subsequence

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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.

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题目：最长递增子串的个数。

思路：动态规划，参考Simple dp solution with explanation。用length[i]和count[i]表示对于子串nums[0...i]最长的的递增子串，以及对应的个数。

对于任意的i，(i > j)，如果nums[i] > nums[j]，有三种情况需要考虑，一种是length[i] < length[j] + 1那么length[i] = length[j] + 1，此时count[i] = count[j]；另一种情况是length[i] == length[j] + 1，那么length[i]不改变，count[i] += count[j]；还有一种情况length[i] > length[j] + 1，不是最长子串跳过。

最后将length[i] == maxlen对应的count[i]求和即可，其中i= 0...n-1。

工程代码下载

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int n = nums.size();
        if(n <= 0) return 0;
        vector<pair<int, int>> dp(n, {1, 1});
        int max_len = 1;
        int res = 0;
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < i; ++j){
                if(nums[j] < nums[i]){
                    if(dp[i].first == dp[j].first + 1)
                        dp[i].second += dp[j].second;
                    else if(dp[i].first < dp[j].first + 1)
                        dp[i] = {dp[j].first + 1, dp[j].second};
                }
            }
            if(dp[i].first == max_len){
                res += dp[i].second;
            }
            else if(dp[i].first > max_len){
                max_len = dp[i].first;
                res = dp[i].second;
            }
        }
        return res;
    }
};