LeetCode – LongestIncreasing Subsequence (Java)

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

For example, given [10, 9, 2, 5, 3, 7, 101, 18], the longestincreasing subsequence is [2, 3, 7, 101]. Therefore the length is 4.

Java Solution 1 - Naive

Letmax[i] represent the length of the longest increasing subsequence so far.
If any element before i is smaller than nums[i], then max[i] = max(max[i],max[j]+1).

Here is an example:

public int lengthOfLIS(int[] nums) {     if(nums==null || nums.length==0)         return 0;       int[] max = new int[nums.length];       for(int i=0; i<nums.length; i++){         max[i]=1;         for(int j=0; j<i;j++){             if(nums[i]>nums[j]){                 max[i]=Math.max(max[i], max[j]+1);             }         }     }       int result = 0;     for(int i=0; i<max.length; i++){         if(max[i]>result)             result = max[i];     }     return result; }

Java Solution 2 - Binary Search

We can put the increasing sequence in a list.

for each num in nums

     if(list.size()==0)

          add num to list

     else if(num > last element in list)

          add num to list

     else 

          replace the element in the list which is the smallest but bigger than num

 

public int lengthOfLIS(int[] nums) {     if(nums==null || nums.length==0)         return 0;       ArrayList<Integer> list = new ArrayList<Integer>();       for(int num: nums){         if(list.size()==0 || num>list.get(list.size()-1)){             list.add(num);         }else{             int i=0;             int j=list.size()-1;               while(i<j){                 int mid = (i+j)/2;                 if(list.get(mid) < num){                     i=mid+1;                 }else{                     j=mid;                 }             }               list.set(j, num);         }     }       return list.size(); }