原题链接在这里:https://leetcode.com/problems/maximum-length-of-pair-chain/description/
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 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].
题解:
按照pair的second number 排序pairs. 再iterate pairs, 若当前pair的second number 小于下个pair的first number, 计数sum++, 否则跳过下个pair.
Note: 用curEnd把当前的second number标记出来, 不要用pair[i][1], 否则 i 跳动时就不是当前pair的second number了.
Time Complexity: O(nlogn). n = pairs.length.
Space: O(1).
AC Java:
1 class Solution { 2 public int findLongestChain(int[][] pairs) { 3 if(pairs == null || pairs.length == 0){ 4 return 0; 5 } 6 7 int len = pairs.length; 8 Arrays.sort(pairs, new Comparator<int []>(){ 9 public int compare(int [] a, int [] b){ 10 return a[1] - b[1]; 11 } 12 }); 13 14 int i = 0; 15 int sum = 0; 16 while(i<len){ 17 sum++; 18 int curEnd = pairs[i][1]; 19 while(i+1<len && curEnd>=pairs[i+1][0]){ 20 i++; 21 } 22 i++; 23 } 24 return sum; 25 } 26 }
时间: 2024-10-14 14:42:23