leetcode646 Maximum Length of Pair Chain java解题代码

题目

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 ifb < 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: 2Explanation: The longest chain is [1,2] -> [3,4]

Note:

1. The number of given pairs will be in the range [1, 1000].

含义，给定一组数组对，类似于[[1,2], [2,3], [3,4]]，如果一个数组元素的第一个数大于另一个元素的第二个数，那么就能构成一个链条，求问给定数组对中最长能组成的链条有几节，一个元素默认算一节

解题思路，这一道题是动态规划题，medium中较低难度的题。我的解题思路是首先将数组按照第一个数进行排序，这里用的是冒泡法，消耗时间会比较多。然后遍历数组，按照第二个数的大小来算的最简单的动态规划，代码如下。

package com;public class Solution {  public static void main(String[] args) throws Exception{    int[][] pairs={{100,101},{2,3},{1,6},{4,8},{9,11},{7,15}};  System.out.println(findLongestChain(pairs));   }     public  static int findLongestChain(int[][] pairs) {          int res=0;          for(int i=0;i<pairs.length;i++){            for(int j=0;j<pairs.length-1;j++)      {       int[] tmp=pairs[0];       if(pairs[j][0]>pairs[j+1][0])       {        tmp=pairs[j+1];        pairs[j+1]=pairs[j];        pairs[j]=tmp;       }             }     }                              int first=pairs[0][0];      int sec=pairs[0][1];      res=1;      for(int i=1;i<pairs.length;i++){                     if(pairs[i][1]<sec)       {        sec=pairs[i][1];               }       else if(pairs[i][0]>sec){        res++;        sec=pairs[i][1];       }                    }                    return res;           }}