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].
 
Analysis:
This problem is actually a LIS question.
Code (Java):
class Solution {
    public int findLongestChain(int[][] pairs) {
        if (pairs == null || pairs.length == 0) {
            return 0;
        }
        
        Arrays.sort(pairs, new PairComparator());
        
        int ans = 1;
        int[] dp = new int[pairs.length];
        
        for (int i = 0; i < pairs.length; i++) {
            dp[i] = 1;
            for (int j = 0; j < i; j++) {
                if (pairs[i][0] > pairs[j][1]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            
            ans = Math.max(ans, dp[i]);
        }
        
        return ans;
    }
}
class PairComparator implements Comparator<int[]> {
    @Override
    public int compare(int[] a, int[] b) {
        if (a[0] != b[0]) {
            return a[0] - b[0];
        }
        return a[1] - b[1];
    }
}
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