646 Maximum Length of Pair Chain

646. Maximum Length of Pair Chain

1. Question

You are givennpairs 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 ifb < 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:

1. The number of given pairs will be in the range [1, 1000].

2. Implementation

(1) Sorting + DP

class Solution {
    public int findLongestChain(int[][] pairs) {
        Arrays.sort(pairs, (a, b)->(a[0] - b[0]));

        int n = pairs.length;
        int[] LIS = new int[n];
        Arrays.fill(LIS, 1);

        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (pairs[j][1] < pairs[i][0] && (LIS[j] + 1 > LIS[i])) {
                    LIS[i] = LIS[j] + 1;
                }
            }
        }

        int res = 1;
        for (int len : LIS) {
            res = Math.max(res, len);
        }
        return res;
    }
}

3. Time & Space Complexity

Sorting + DP:时间复杂度O(nlogn + n^2), n为pairs的个数, 空间复杂度O(n)