标签：targe [] div end bsp put height search span
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]).
1 public class Solution { 2 public int maxEnvelopes(int[][] envelopes) { 3 if(envelopes==null||envelopes.length==0||envelopes[0]==null||envelopes[0].length!=2) return 0; 4 Arrays.sort(envelopes,new Comparator<int[]>(){ 5 public int compare(int[] i1,int[] i2){ 6 if(i1[0]==i2[0]) return i2[1]-i1[1]; 7 else return i1[0]-i2[0]; 8 } 9 }); 10 int[] dp = new int[envelopes.length]; 11 int len = 0; 12 for(int[] envelope:envelopes){ 13 int end =binarysearch(dp,0,len,envelope[1]); 14 if(end==len) len++; 15 } 16 return len; 17 } 18 public int binarysearch(int[] dp,int left,int right,int target){ 19 while(left<right){ 20 int mid = left+(right-left)/2; 21 if(dp[mid]>=target) right = mid; 22 else left = mid+1; 23 } 24 dp[left] = target; 25 return left; 26 } 27 } 28 //the time complexity could be nlogn, the space complexity could be O(n);
标签：targe [] div end bsp put height search span
原文地址：http://www.cnblogs.com/codeskiller/p/6939599.html
