Russian Doll Envelopes

最新推荐文章于 2020-04-23 10:54:26 发布

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#java #leetcode #动态规划

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本文探讨了俄罗斯套娃信封问题的算法解决方案，通过宽度升序高度降序的排序方式，结合动态规划思想，有效地提高了寻找最大套娃信封序列的效率。

题目描述：

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]).

自己写了一个回溯的算法，先排序，然后计算从envelopes[0]为起点，能装入的最大包裹数，然后从envelopes[1]为起点，以此类推。遗憾的是超时了，代码如下：

public class Solution {
    public int maxEnvelopes(int[][] envelopes) {
        int n=envelopes.length;
        if(n==0)
        	return 0;
        Arrays.sort(envelopes, new Comparator<int[]>(){
            public int compare(int[] arr1, int[] arr2){
                if(arr1[0] == arr2[0])
                    return arr1[1] - arr2[1];
                else
                    return arr1[0] - arr2[0];
           } 
        });
		int max=0;
		boolean[] started=new boolean[envelopes.length];
		for(int i=0;i<n;i++){
			if(!started[i]){
				max=max>getMaxEnvelopes(envelopes, i, started)?max:getMaxEnvelopes(envelopes, i, started);
			}
		}
		return max;
    }
	
	public int getMaxEnvelopes(int[][] envelopes,int startIndex,boolean[] started){
		int max=1;
		int width=envelopes[startIndex][0];
		int height=envelopes[startIndex][1];
		for(int i=startIndex+1;i<envelopes.length;i++){
			if(envelopes[i][0]>width&&envelopes[i][1]>height){
				started[i]=true;
				int num=getMaxEnvelopes(envelopes, i , started)+1;
				max=max>num?max:num;
			}
		}
		return max;
	}
}

大神的解法，width升序，height降序！：

public class Solution {
    public int maxEnvelopes(int[][] envelopes) {
	    if(envelopes == null || envelopes.length == 0 
	       || envelopes[0] == null || envelopes[0].length != 2)
	        return 0;
	    Arrays.sort(envelopes, new Comparator<int[]>(){
	        public int compare(int[] arr1, int[] arr2){
	            if(arr1[0] == arr2[0])
	                return arr2[1] - arr1[1];
	            else
	                return arr1[0] - arr2[0];
	       } 
	    });
	    int dp[] = new int[envelopes.length];
	    int len = 0;
	    for(int[] envelope : envelopes){
	        int index = Arrays.binarySearch(dp, 0, len, envelope[1]);
	        if(index < 0)
	            index = -(index + 1);
	        dp[index] = envelope[1];
	        if(index == len)
	            len++;
	    }
	    return len;
	}
}