646. Maximum Length of Pair Chain

ou 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].

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/maximum-length-of-pair-chain
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动态规划：

对数组按照结尾的数字进行排序，dp[i]表示第i个区间能放的最多的数字。

class Solution {
    public int findLongestChain(int[][] pairs) {
        if (pairs.length == 0) return 0;
        int ans = 1;
        Arrays.sort(pairs, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[1] - o2[1];
            }
        });
        int []dp = new int[pairs.length];
        Arrays.fill(dp, 1);
        for (int i = 1; i < pairs.length; i++) {
            for (int j = i - 1; j >= 0; j--) {
                if (pairs[j][1] < pairs[i][0]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                    ans = Math.max(ans, dp[i]);
                }
            }
        }
        return ans;
    }
}

二，贪心算法

class Solution {
    public int findLongestChain(int[][] pairs) {
        if (pairs.length == 0) return 0;
        int ans = 1;
        Arrays.sort(pairs, new Comparator<int[]>() {
            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[1] - o2[1];
            }
        });
        int tmpEnd = pairs[0][1];
        for (int i = 1; i < pairs.length; i++) {
            if (pairs[i][0] > tmpEnd) {
                ans++;
                tmpEnd = pairs[i][1];
            }
        }
        return ans;
    }
}