README_EN.md

300. Longest Increasing Subsequence

中文文档

Description

Given an integer array nums, return the length of the longest strictly increasing subsequence.

A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].

 

Example 1:

Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.

Example 2:

Input: nums = [0,1,0,3,2,3]
Output: 4

Example 3:

Input: nums = [7,7,7,7,7,7,7]
Output: 1

 

Constraints:

• 1 <= nums.length <= 2500
• -104 <= nums[i] <= 104

 

Follow up:

• Could you come up with the O(n2) solution?
• Could you improve it to O(n log(n)) time complexity?

Solutions

Dynamic programming.

Python3

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [1] * n
        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

Java

class Solution {
    public int lengthOfLIS(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        int res = 1;
        for (int i = 1; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                if (nums[j] < nums[i]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            res = Math.max(res, dp[i]);
        }
        return res;
    }
}

TypeScript

function lengthOfLIS(nums: number[]): number {
    let n = nums.length;
    let dp = new Array(n).fill(1);
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
    }
    return Math.max(...dp);
};

C++

class Solution {
public:
    int lengthOfLIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n, 1);
        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j < i; ++j)
            {
                if (nums[j] < nums[i]) dp[i] = max(dp[i], dp[j] + 1);
            }
        }
        return *max_element(dp.begin(), dp.end());
    }
};

Go

func lengthOfLIS(nums []int) int {
	n := len(nums)
	dp := make([]int, n)
	dp[0] = 1
	res := 1
	for i := 1; i < n; i++ {
		dp[i] = 1
		for j := 0; j < i; j++ {
			if nums[j] < nums[i] {
				dp[i] = max(dp[i], dp[j]+1)
			}
		}
		res = max(res, dp[i])
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

...