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: 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.

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1
int 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;
}