[Likou] 300. Longest Increasing Subsequence＜Dynamic Programming＞

[Likou] 300. Longest increasing subsequence

Given an array of integers nums, find the length of the longest strictly increasing subsequence in it.
A subsequence is a sequence derived from an array by removing (or not removing) elements from the array 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], so 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

提示：
1 <= nums.length <= 2500
-$10^4$ <= nums[i] <= $10^4$

answer

Retrace five steps:

• Determine the meaning of the dp array and the subscript, which
  dp[j] means: iThe length of the longest ascending subsequence ending with the th number. Note nums[i]that must be selected .

• Determine the recurrence formula
  

• How to initialize dp array
  dp[0]=1

• Determine the order of traversal.
  Normal traversal is from left to right.

• Take an example to deduce the dp array.
  The input is: 10 9 2 5 3 7 101, and the output is: 1 1 1 2 2 3 4
  The input is: 5 7 1 9 4 6 2 8 3, and the output is:1 2 1 3 2 3 2 4 3

class Solution {
    
    
    public int lengthOfLIS(int[] nums) {
    
    
        if (nums.length == 0) {
    
    
            return 0;
        }
        
        int[] dp = new int[nums.length];
        dp[0] = 1;
        int maxans = 1;
        
        for (int i = 1; i < nums.length; i++) {
    
    
        	// 默认以i为结尾的最长串是1，只有它本身
            dp[i] = 1;
            // 从0 判断到 i
            for (int j = 0; j < i; j++) {
    
    
                if (nums[i] > nums[j]) {
    
    
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            // 取dp数组中的最大值
            maxans = Math.max(maxans, dp[i]);
        }
        return maxans;
    }
}

When calculating state dp[i], it takes O(n) time to traverse all states of dp[0…i−1], so the total time complexity is O( n $n^2$)