Leetcode/微软-- 300. Longest Increasing Subsequence(DP)

300. Longest Increasing Subsequence(DP)

Given an unsorted array of integers, find the length of longest increasing subsequence.

For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.

Your algorithm should run in O(n²) complexity.

思路：double loop, j is element before i, dp[i] is current element seq max length eg[3,4,-1,0,6,2,3]

arr[j] < arr[i]: dp[i] is Max(dp[j]+1,dp[i])
                  //not update i if i already bigger(results from previous: 3,4,6; so not update when get to -1)
else not increaing: j++
once j=i: i++,j=0

Complexity: O(n^2)

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