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].
[Thoughts]
按照数组的结束数字排序,然后从左到右扫描。如果满足b<c的条件,则count+1.
[Code]
1: int findLongestChain(vector<vector<int>>& pairs) {
2: if(pairs.size() == 0) return 0;
3:
4: std::sort(pairs.begin(), pairs.end(), comp);
5:
6: int count = 1;
7: vector<int>& pair = pairs[0];
8: for(int i =1; i<pairs.size(); i++) {
9: if(pairs[i][0] > pair[1]) {
10: count ++;
11: pair = pairs[i];
12: }
13: }
14:
15: return count;
16: }
17:
18: static bool comp(vector<int>& a, vector<int>& b) {
19: return a[1] < b[1];
20: }
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