這道題麼，找最長的鏈，所謂的鏈就是上一個的結尾一定大於下一個的開頭

因為要找最長的鏈，所以我們按照結尾的那個數字排序，這樣可以儘量的最長
然後從第一個開始，直接往後遍歷找就好，反正在前面的一定是一個最優解（因為不要求區間覆蓋的覆蓋長度，而是要求pair的數量）

相對的，如果是要找區間最長，應該就是排序後要上DP了，而不是這樣直接找

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.

Now, we define a pair (c, d) can follow another pair (a, b) if
 
 and only if b < c. Chain of pairs can be formed in this fashion.

Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.

Example 1:
Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is
 
 [1,2] -> [3,4]
Note:
The number of given pairs will be in the range [1, 1000].

public class Solution {
   public int findLongestChain(int[][] pairs) {
       //sort
       /**
       按照截止排序，就可以找到最優的組合方案了
       **/
       Arrays.sort(pairs, (a,b) -> a[1] - b[1]);
       int sum = 0, n = pairs.length;
       int
 
 i = 0;
       while (i < n){
           int current_end = pairs[i][1];
           sum ++;
           while ( i < n && pairs[i][0] <= current_end) i++;
       }
       return sum;

    }
}