Content

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

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: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

Solution

1. 动态规划

Java

class Solution {
    public int lengthOfLIS(int[] nums) {
        // 1 <= nums.length <= 2500
        // -10⁴ <= nums[i] <= 10⁴
        int n = nums.length;
        // dp[i]表示以下标为i的元素结尾的最长子序列的长度
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        int max = 1;
        for (int i = 1; i < n; i++) {
            int j = i - 1;
            // 遍历i之前的所有数字，
            // 1. 当nums[j] < nums[i]时，nums[i]可以添加到dp[j]所代表的子序列之后。
            // 2. 当nums[j] == nums[i]时，nums[i]可以替换调dp[j]所代表的子序列的最后一个数字即nums[j]。
            // 3. 当num[j] > nums[j]时，跳过。
            while (j >= 0) {
                if (nums[j] < nums[i]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                } else if (nums[j] == nums[i]) {
                    dp[i] = Math.max(dp[i], dp[j]);
                }
                j--;
            }
            max = Math.max(max, dp[i]);
        }
        return max;
    }
}