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.

Java:

public int findNumberOfLIS(int[] nums) {
int n = nums.length, res = 0, max_len = 0;
int[] len =  new int
, cnt = new int
;
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;
}　　

Python:

class Solution(object):
def findNumberOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
# Time: O(n^2)
# Space: O(n)
dp, longest = [[1, 1] for i in range(len(nums))], 1
for i, num in enumerate(nums):
curr_longest, count = 1, 0
for j in range(i):
if nums[j] < num:
curr_longest = max(curr_longest, dp[j][0] + 1)
for j in range(i):
if dp[j][0] == curr_longest - 1 and nums[j] < num:
count += dp[j][1]
dp[i] = [curr_longest, max(count, dp[i][1])]
longest = max(curr_longest, longest)
return sum([item[1] for item in dp if item[0] == longest])

Python:

class Solution(object):
def findNumberOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
dp = [[1, 1] for i in range(len(nums))]
max_for_all = 1
for i, num in enumerate(nums):
max_len, count = 1, 0
for j in range(i):
if nums[j] < num:
if dp[j][0] + 1 > max_len:
max_len = dp[j][0] + 1
count = 0
if dp[j][0] == max_len - 1:
count += dp[j][1]
dp[i] = [max_len, max(count, dp[i][1])]
max_for_all = max(max_len, max_for_all)
return sum([item[1] for item in dp if item[0] == max_for_all])

Python: wo

class Solution(object):
def findNumberOfLIS(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
if not nums:
return 0

n = len(nums)
res = 0
cnt = [1] * n
mx = [1] * n
max_len = 1
for i in xrange(n):
cur_longest = 1
for j in xrange(i):
temp = 1
if nums[i] > nums[j]:
temp = mx[j] + 1
if cur_longest < temp:
cur_longest = temp
mx[i] = cur_longest
cnt[i] = cnt[j]
elif cur_longest == temp:
cnt[i] += cnt[j]
if mx[i] > max_len:
max_len = mx[i]
res = cnt[i]
elif mx[i] == max_len:
res += cnt[i]

return res     　　

C++:

int findNumberOfLIS(vector<int>& nums) {
int n = nums.size(), res = 0, max_len = 0;
vector<pair<int,int>> dp(n,{1,1});            //dp[i]: {length, number of LIS which ends with nums[i]}
for(int i = 0; i<n; i++){
for(int j = 0; j <i ; j++){
if(nums[i] > nums[j]){
if(dp[i].first == dp[j].first + 1)dp[i].second += dp[j].second;
if(dp[i].first < dp[j].first + 1)dp[i] = {dp[j].first + 1, dp[j].second};
}
}
if(max_len == dp[i].first)res += dp[i].second;
if(max_len < dp[i].first){
max_len = dp[i].first;
res = dp[i].second;
}
}
return res;
}

C++:　　

class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
int res = 0, mx = 0, n = nums.size();
vector<int> len(n, 1), cnt(n, 1);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[i] <= nums[j]) continue;
if (len[i] == len[j] + 1) cnt[i] += cnt[j];
else if (len[i] < len[j] + 1) {
len[i] = len[j] + 1;
cnt[i] = cnt[j];
}
}
if (mx == len[i]) res += cnt[i];
else if (mx < len[i]) {
mx = len[i];
res = cnt[i];
}
}
return res;
}
};

C++:　　

class Solution {
public:
int findNumberOfLIS(vector<int>& nums) {
int res = 0, mx = 0, n = nums.size();
vector<int> len(n, 1), cnt(n, 1);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < i; ++j) {
if (nums[i] <= nums[j]) continue;
if (len[i] == len[j] + 1) cnt[i] += cnt[j];
else if (len[i] < len[j] + 1) {
len[i] = len[j] + 1;
cnt[i] = cnt[j];
}
}
mx = max(mx, len[i]);
}
for (int i = 0; i < n; ++i) {
if (mx == len[i]) res += cnt[i];
}
return res;
}
};

类似题目：

[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列

[LeetCode] 674. Longest Continuous Increasing Subsequence 最长连续递增序列

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