[leetcode]354. Russian Doll Envelopes

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and
height of one envelope are greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Example: Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]).

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这题就是一个二维的LIS(longest increasing subsequence)。宽度进行递增排序，高度进行递减排序。
对于递增序列，当前值如果比尾部的值大则增加，在中间的某个值，则更新（即将值替换成一个较小的，增加长度变长的可能）。
Arrays.sort(envelope, new Comparator<int[]>(){}) envelope是一个二维数组，里面的一维数组是被比较的部分。

public class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        if(envelopes == null || envelopes.length == 0) return 0;
        Arrays.sort(envelopes, new Comparator<int[]>(){
           @Override
           public int compare(int[] arr1, int[] arr2){
               if(arr1[0] == arr2[0]){
                  return arr2[1] - arr1[1];
               } else{
                  return arr1[0] - arr2[0];
               }
           }
        });
        
        int[] dp = new int[envelopes.length];   // the tail nums of length i;
        int len = 0;
        for(int[] envelope: envelopes){
            int index = Arrays.binarySearch(dp, 0, len, envelope[1]);
            if(index < 0){                  // do not contain this num, adding to tail
                index = -(index + 1);       // replace it with smaller one
            }
            dp[index] = envelope[1];
            if(index == len) len++;
        }
        return len; 
    }