354. Russian Doll Envelopes

最新推荐文章于 2019-01-18 10:09:35 发布

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本文探讨了俄罗斯套娃信封问题的解决方案，即如何找到能够互相套叠的最大数量信封组。首先按宽度升序排列，当宽度相等时按高度降序排列。之后利用最长递增子序列算法（LIS）来确定最大套叠数量。

You have a number of envelopes with widths and heights given as a pair of integers (w, h). One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and height of the other envelope.

What is the maximum number of envelopes can you Russian doll? (put one inside other)

Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]], the maximum number of envelopes you can Russian doll is 3([2,3] => [5,4] => [6,7]).

先通过w按照升序排序，在按照h的升序排序（当w相同时，h降序），再结合LIS求解，程序如下所示：


class Solution {
    class Node {
        int weight;
        int height;
        Node(int w, int h){
            weight = w;
            height = h;
        }
    }
    List<Node> list = new ArrayList<>();
    public int maxEnvelopes(int[][] envelopes) {
        if (envelopes.length == 0){
            return 0;
        }
        for (int i = 0; i < envelopes.length; ++ i){
            list.add(new Node(envelopes[i][0], envelopes[i][1]));   
        }
        Collections.sort(list, new Comparator<Node>(){
            @Override
            public int compare(Node m, Node n){
                if (m.weight != n.weight){
                    return m.weight - n.weight;
                }
                return n.height - m.height;
            }
        }); 
        
        int len = list.size();
        int[] heightArr = new int[len];
        for (int i = 0; i < len; ++ i){
            heightArr[i] = list.get(i).height;
        }
        return LIS(heightArr);
    }
    
    public int LIS(int[] height){
        int len = height.length;
        if (len <= 1){
            return len;
        }
        int[] array = new int[len];
        int maxVal = 0;
        for (int i = 1; i < len; ++ i){
            for (int j = 0; j < i; ++ j){
                if (height[i] > height[j]){
                    array[i] = Math.max(array[i], array[j] + 1);
                }
            }
            maxVal = Math.max(maxVal, array[i]);
        }
        return maxVal + 1;
    }
}