[LeetCode] 354. Russian Doll Envelopes 俄罗斯套娃信封

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)

Note:
Rotation is not allowed.

Example:

Input: [[5,4],[6,4],[6,7],[2,3]]
Output: 3 Explanation: The maximum number of envelopes you can Russian doll is  ([2,3] => [5,4] => [6,7]).3

信封的嵌套问题，就像俄罗斯套娃同样，一个套一个，求能套的最多数量。

和300. Longest Increasing Subsequence相似，从那题一维变成了两维。

先给信封排序，按信封的宽度从小到大排，宽度相等时，高度大的在前面。问题就简化了成了找高度数字中的Longest Increasing Subsequence。

Java: Naive

public int maxEnvelopes(int[][] envelopes) {
    if(envelopes==null||envelopes.length==0)
        return 0;
 
    Arrays.sort(envelopes, new Comparator<int[]>(){
        public int compare(int[] a, int[] b){
            if(a[0]!=b[0]){
                return a[0]-b[0];
            }else{
                return a[1]-b[1];
            }
        }
    });
    int max=1;
    int[] arr = new int[envelopes.length];
    for(int i=0; i<envelopes.length; i++){
        arr[i]=1;
        for(int j=i-1; j>=0; j--){
            if(envelopes[i][0]>envelopes[j][0]&&envelopes[i][1]>envelopes[j][1]){
                arr[i]=Math.max(arr[i], arr[j]+1);
            }
        }
        max = Math.max(max, arr[i]);
    }
 
    return max;
}　

Java: Binary Search

public int maxEnvelopes(int[][] envelopes) {
    if(envelopes==null||envelopes.length==0)
        return 0;
 
    Arrays.sort(envelopes, new Comparator<int[]>(){
        public int compare(int[] a, int[] b){
            if(a[0]!=b[0]){
                return a[0]-b[0]; //ascending order
            }else{
                return b[1]-a[1]; // descending order
            }
        }
    });
 
    ArrayList<Integer> list = new ArrayList<Integer>();
 
    for(int i=0; i<envelopes.length; i++){
 
        if(list.size()==0 || list.get(list.size()-1)<envelopes[i][1])
            list.add(envelopes[i][1]);
 
        int l=0;
        int r=list.size()-1;
 
        while(l<r){
            int m=l+(r-l)/2;
            if(list.get(m)<envelopes[i][1]){
                l=m+1;
            }else{
                r=m;
            }
        }
 
        list.set(r, envelopes[i][1]);
    }
 
    return list.size();
}

Python：

class Solution(object):
    def maxEnvelopes(self, envelopes):

        def insert(target):
            left, right = 0, len(result) - 1
            while left <= right:
                mid = left + (right - left) / 2
                if result[mid] >= target:
                    right = mid - 1
                else:
                    left = mid + 1
            if left == len(result):
                result.append(target)
            else:
                result[left] = target

        result = []
        envelopes.sort(lambda x, y: y[1] - x[1] if x[0] == y[0] else \
                                    x[0] - y[0])
        for envelope in envelopes:
            insert(envelope[1])

        return len(result)

C++：

class Solution {
public:
    int maxEnvelopes(vector<pair<int, int>>& envelopes) {
        vector<int> dp;
        sort(envelopes.begin(), envelopes.end(), [](const pair<int, int> &a, const pair<int, int> &b){
            if (a.first == b.first) return a.second > b.second;
            return a.first < b.first;
        });
        for (int i = 0; i < envelopes.size(); ++i) {
            int left = 0, right = dp.size(), t= envelopes[i].second;
            while (left < right) {
                int mid = left + (right - left) / 2;
                if (dp[mid] < t) left = mid + 1;
                else right = mid;
            }
            if (right >= dp.size()) dp.push_back(t);
            else dp[right] = t;
        }
        return dp.size();
    }
};

 

相似题目：

[LeetCode] 300. Longest Increasing Subsequence 最长递增子序列

 

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