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].
思路:
动态规划。递推公式是 dp[i] = max(dp[i],dp[j]+1);其中0=<j<i;
int findLongestChain(vector<vector<int>>& pairs) { sort(pairs.begin(), pairs.end(), [](const vector<int>&a, const vector<int>&b){return a[0] < b[0]; }); vector<int> dp(pairs.size(), 1); for (int i = 1; i < pairs.size(); i++) { for (int j = 0; j<i;j++) { if (pairs[i][0]>pairs[j][1]) { dp[i] = max(dp[i],dp[j]+1); } } } return *max_element(dp.begin(), dp.end()); }
时间: 2024-10-08 04:59:48