Lintcode-medium-longest increasing subsequence

Given a sequence of integers, find the longest increasing subsequence (LIS).

You code should return the length of the LIS.

Clarification

What ' s the definition of longest increasing subsequence?

* The longest increasing subsequence problem is to find a subsequence of a given sequence in which the subsequence ' s Eleme NTS is in sorted order, lowest to highest, and in which the subsequence is as long as possible. This subsequence are not necessarily contiguous, or unique.

* Https://en.wikipedia.org/wiki/Longest_common_subsequence_problem

Example

For [5, 4, 1, 2, 3], the LIS is [1, 2, 3], return 3

For [4, 2, 4, 5, 3, 7], the LIS is [4, 4, 5, 7], return 4

Dynamic planning:

Use an int array to represent the number of I as the last subsequence length

So as long as the back number is larger than the previous number, the oldest sequence length is updated.

Finally, traversing one side of the DP array, select the maximum value

 Public classSolution {/**     * @paramnums:the Integer Array *@return: The length of LIS (longest increasing subsequence)*/     Public intLongestincreasingsubsequence (int[] nums) {        //Write your code here                if(Nums = =NULL|| Nums.length = = 0)            return0; int[] DP =New int[Nums.length]; intLongest = 1;  for(inti = 0; i < nums.length; i++) Dp[i]= 1;  for(inti = 0; i < dp.length; i++){             for(intj = i + 1; J < Dp.length; J + +){                if(Nums[j] >=Nums[i]) {Dp[j]= Math.max (Dp[j], dp[i] + 1); }            }        }                 for(inti = 0; i < dp.length; i++) Longest=Math.max (Longest, dp[i]); returnlongest; }}

Lintcode-medium-longest increasing subsequence