[LeetCode] 673. Number of Longest Increasing Subsequence 最长递增序列的个数

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[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;
    }　　

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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转载于:https://www.cnblogs.com/lightwindy/p/9721726.html