646. Maximum Length of Pair Chain（动态规划 vs 贪心）

1.Description

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].

---

2.Analysis

其实这道题的解法更像是贪心，先对pairs排序，排序根据pair对的second变量递增排序。再进行遍历，当新的pairs <ai+1,bi+1>，如果前面存在<ai,bi>，在ai+1不小于ai的情况下，可以将pair加入，一旦小于则会使当前的pair链规模减小，达不到最大的要求。

---

3.Code

C++ 贪心

class Solution {public:    static bool compare(const vector<int>& a, const vector<int>& b) {        return a[1] < b[1]? true :false;    }    int findLongestChain(vector<vector<int> >& pairs) {       if(pairs.size() == 0) return 0;       vector<vector<int> > res;       //根据second对pairs排序        sort(pairs.begin(),pairs.end(), compare);       vector<int> min = pairs[0];       res.push_back(min);       int count = 1;       for(int i = 1; i < pairs.size(); i++) {       //在不减少结果规模时插入向量尾            if(pairs[i][0] > min[1]) {                res.push_back(pairs[i]);                min = pairs[i];                count++;            }       }       return count;    }};

Java 动态规划

    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];    }