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].
dp的思想
class Solution {
public:
int findLongestChain(vector<vector<int>>& pairs) {
int len = pairs.size();
if (len == ){
return ;
}
int res = ;
sort(pairs.begin(), pairs.end(), cmp);
vector<int> dp(len, );
for (int i = ; i < len; i++){
for (int j = ; j < i; j++){
if (pairs[i][] > pairs[j][]){
dp[i] = max(dp[i], dp[j] + );
}
}
}
return dp[len - ];
}
private:
static bool cmp(vector<int>& a, vector<int>&b){
return a[] < b[];
}
};