leetcode 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.

 

题目大意：求最长递增子序列的数量。

 

方法一：动态规划

class Solution {
public:
    int findNumberOfLIS(vector<int>& nums) {
        int len = nums.size();
        vector<int> dp(len, 1); // dp[i] = length of longest ending in nums[i]
        vector<int> count(len, 1); // count[i] = number of longest ending in nums[i]
        int maxn = 0;
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    if (dp[j] >= dp[i]) {
                        dp[i] = dp[j] + 1;
                        count[i] = count[j];
                    } else {
                        if (dp[j] + 1 == dp[i])
                            count[i] += count[j];
                    }
                }
            }
            maxn = max(maxn, dp[i]);
        }
        int cnt = 0;
        for (int i = 0; i < len; i++) {
            if (dp[i] == maxn)
                cnt += count[i];
        }
        return cnt;
    }
};

方法二：线段树

https://leetcode.com/articles/a-recursive-approach-to-segment-trees-range-sum-queries-lazy-propagation/