DP20 求pair的最大长度链 Maximum Length Chain of Pairs @geeksforgeeks

最新推荐文章于 2022-01-12 00:12:38 发布

最新推荐文章于 2022-01-12 00:12:38 发布 · 138 阅读

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#java #数据结构与算法

本文介绍如何通过排序和动态规划算法解决给定一对数的最长链问题，以亚马逊面试题为例，详细阐述了解题步骤和代码实现。

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是LIS的变型题，如果遇到两维则先排序一维再对另一维动态规划

You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.A pair (c, d) can follow another pair (a, b) if b < c. Chain of pairs can be formed in this fashion. Find the longest chain which can be formed from a given set of pairs.
Source:Amazon Interview | Set 2

For example, if the given pairs are {{5, 24}, {39, 60}, {15, 28}, {27, 40}, {50, 90} }, then the longest chain that can be formed is of length 3, and the chain is {{5, 24}, {27, 40}, {50, 90}}

This problem is a variation of standardLongest Increasing Subsequenceproblem. Following is a simple two step process.
1) Sort given pairs in increasing order of first (or smaller) element.
2) Now run a modified LIS process where we compare the second element of already finalized LIS with the first element of new LIS being constructed.

The following code is a slight modification of method 2 ofthis post.

package DP;

import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;

public class MaximumLengthChainOfPairs {

	public static void main(String[] args) {
		Pair p1 = new Pair(5, 24);
		Pair p2 = new Pair(39, 60);
		Pair p3 = new Pair(15, 28);
		Pair p4 = new Pair(27, 40);
		Pair p5 = new Pair(50, 90);
		Pair[] A = {p1,p2,p3,p4,p5};
		int n = A.length;
		System.out.println(maxChainLength(A, n));	// {{5, 24}, {27, 40}, {50, 90}}
	}
	
	public static int maxChainLength(Pair[] A, int n){
		Arrays.sort(A, new Comparator<Pair>() {
			@Override
			public int compare(Pair o1, Pair o2) {
				return o1.a - o2.a;
			}
		});
		
//		System.out.println(Arrays.toString(A));
		
		int[] mcl = new int[n];
		int max = 0;
		
		/* Initialize MCL (max chain length) values for all indexes */
		for(int i=0; i<n; i++){
			mcl[i] = 1;
		}
		
		/* Compute optimized chain length values in bottom up manner */
		for(int i=1; i<n; i++){
			for(int j=0; j<i; j++){
				if(A[j].b < A[i].a){		// 保证上升
					mcl[i] = Math.max(mcl[i], mcl[j]+1);
				}
			}
		}
		// mcl[i] now stores the maximum chain length ending with pair i
		 /* Pick maximum of all MCL values */
		for(int i=0; i<n; i++){
			max = Math.max(max, mcl[i]);
		}
		
		return max;
	}

	
	public static class Pair{
		int a;
		int b;
		public Pair(int a_, int b_){
			a = a_;
			b = b_;
		}
		@Override
		public String toString() {
			return "Pair [a=" + a + ", b=" + b + "]";
		}
		
	}
}

http://www.geeksforgeeks.org/dynamic-programming-set-20-maximum-length-chain-of-pairs/