[DP] Maximum non-overlapping Intervals

1. Example
Find the maximum number of non-overlapping intervals
s= [[0,2], [1,4], [3,5]]
the result is 2


2. Implementation
dp[i] represents the max number of non-overlapping intervals for 0~ i with i selected
non-overlapping => a.end =< b.start && b.end > b.start

collections.sort(intervals, new EndComparator(){
     @ Override
     public int compare (interval a, interval b)
     {
          return a.end - b.end;
     }
});

int[] dp = new int[intervals.size()];
int max = 0;
for (int i = 1 ; i < intervals.size(); i ++ )
{
    for ( int j = 1-1; j >=0 ; j--)
    {
          if (intervals[j].end <= intervals[i].start)
              dp[i] = Math.max( dp[j]+1, dp[i] );
    }
    max = Math.max(max, dp[i]);
}

return max; 

3. Similar Ones
http://sidbai.github.io/2015/07/07/Maximum-Non-overlapping-Intervals/