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.
The idea is to use two arrays len[n] and cnt[n] to record the maximum length of Increasing Subsequence and the coresponding number of these sequence which ends with nums[i], respectively. That is:
len[i]: the length of the Longest Increasing Subsequence which ends with nums[i].cnt[i]: the number of the Longest Increasing Subsequence which ends with nums[i].
Then, the result is the sum of each cnt[i] while its corresponding len[i] is the maximum length.
Java version:
public int findNumberOfLIS(int[] nums) {
int n = nums.length, res = 0, max_len = 0;
int[] len = new int[n], cnt = new int[n];
for(int i = 0; i<n; i++){
len[i] = cnt[i] = 1;
for(int j = 0; j <i ; j++){
if(nums[i] > nums[j]){
if(len[i] == len[j] + 1)cnt[i] += cnt[j];
if(len[i] < len[j] + 1){
len[i] = len[j] + 1;
cnt[i] = cnt[j];
}
}
}
if(max_len == len[i])res += cnt[i];
if(max_len < len[i]){
max_len = len[i];
res = cnt[i];
}
}
return res;
}
