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.
2. Implementation
(1) DP
class Solution {
public int findNumberOfLIS(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int n = nums.length;
// len[i] is the length of longest subsequence that end with nums[i]
int[] len = new int[n];
// count[i] is the number of longest subsequence that end with nums[i]
int[] count = new int[n];
Arrays.fill(len, 1);
Arrays.fill(count, 1);
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i]) {
if (len[j] + 1 == len[i]) {
count[i] += count[j];
}
else if (len[j] + 1 > len[i]) {
len[i] = len[j] + 1;
count[i] = count[j];
}
}
}
}
int maxLen = 1;
for (int l : len) {
maxLen = Math.max(maxLen, l);
}
int res = 0;
for (int i = 0; i < n; i++) {
if (maxLen == len[i]) {
res += count[i];
}
}
return res;
}
}