673[Medium]: Number of Longest Increasing Subsequence

Part1:问题描述

Given an unsorted array of integers, find the number of longest increasing subsequence.

Example 1:

Input: [1,3,5,4,7]Output: 2Explanation: The two longest increasing subsequence are [1, 3, 4, 7] and [1, 3, 5, 7].

Example 2:

Input: [2,2,2,2,2]Output: 5Explanation: 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.

Part2:解题思路

Part3:代码

#include<iostream>#include<vector>#include<algorithm>using namespace std;int findNumberOfLIS(vector<int>& nums) {int result = 0;// 注意最小的长度是1int maxlen = 1;int length = nums.size();if (length == 0) return result; // len[i]以nums[i]结尾的最长递增子序列的长度 int* len = new int[length + 1];for (int i = 0; i < length+1; i++) {len[i] = 1;}// count[i]以nums[i]结尾的最长递增子序列的个数 int* count = new int[length + 1];for (int i = 0; i < length+1; i++) {count[i] = 1;}for (int i = 0; i < length; i++) {for (int j = 0; j < i; j++) {if (i == 0) {len[i] = 1;count[i] = 1;break;} else if (nums[i] > nums[j]) {if (len[j] + 1 > len[i]) {len[i] = len[j] + 1;count[i] = count[j];} else if (len[j] + 1 == len[i]) {count[i] += count[j];}}maxlen = max(maxlen, len[i]);}}for (int i = 0; i < length; i++) {if (len[i] == maxlen) {result += count[i];}}return result;}int main() {int num;cin >> num;vector<int> nums;for (int i = 0; i < num; i++) {int temp;cin >> temp;nums.push_back(temp);}cout << findNumberOfLIS(nums) << endl;return 0;}