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].
- Approach #1: Greedy. [C++]
- class Solution {
- public:
- int findLongestChain(vector<vector<int>>& pairs) {
- sort(pairs.begin(), pairs.end(), cmp);
- int ans = 1;
- int idx = 0;
- for (int i = 1; i <pairs.size(); ++i) {
- if (pairs[i][0]> pairs[idx][1]) {
- ans++;
- idx = i;
- }
- }
- return ans;
- }
- static bool cmp(vector<int> a, vector<int> b) {
- return a[1] <b[1];
- }
- };
- Approach #2: DP. [Java]
- class Solution {
- public int findLongestChain(int[][] pairs) {
- if (pairs == null || pairs.length == 0) return 0;
- Arrays.sort(pairs, (a, b) -> (a[0] - b[0]));
- int[] dp = new int[pairs.length];
- Arrays.fill(dp, 1);
- for (int i = 0; i <dp.length; ++i) {
- for (int j = 0; j < i; ++j) {
- dp[i] = Math.max(dp[i], pairs[i][0]> pairs[j][1] ? dp[j] + 1: dp[j]);
- }
- }
- return dp[pairs.length - 1];
- }
- }
来源: http://www.bubuko.com/infodetail-2983941.html