Description
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.
Solution
DP, time O(n ^ 2), space O(n)
是"300. Longest Increasing Subsequence"的简单扩展。
class Solution {
public int findNumberOfLIS(int[] nums) {
int n = nums.length;
int[] dpLen = new int[n];
int[] dpCount = new int[n];
int maxLen = 0;
int maxCount = 0;
for (int i = 0; i < n; ++i) {
dpLen[i] = 1;
dpCount[i] = 1;
for (int j = 0; j < i; ++j) {
if (nums[i] <= nums[j]) {
continue;
}
if (dpLen[i] < dpLen[j] + 1) {
dpLen[i] = dpLen[j] + 1;
dpCount[i] = dpCount[j];
} else if (dpLen[i] == dpLen[j] + 1) {
dpCount[i] += dpCount[j];
}
}
if (dpLen[i] > maxLen) {
maxLen = dpLen[i];
maxCount = dpCount[i];
} else if (dpLen[i] == maxLen) {
maxCount += dpCount[i];
}
}
return maxCount;
}
}
