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)
解题思路:
这道题个人觉得没有别的hard难度的题复杂。难点可能在于对数据的操作。为此我新建一个”信封“类,把二维数组的每一行,即每一个“信封”封装成了一个对象,存在数组中。还新建了一个MyCompatator类方便为信封数组进行排序。经过这样的抽象化步骤,这个问题就变成了很简单的有序数组中二次循环找最大值的问题。这就十分容易解决了。
具体代码:
1 public class Solution { 2 3 public static int maxEnvelopes(int[][] envelopes) { 4 //特殊情况的判断 5 if(envelopes==null||envelopes.length==0) 6 return 0; 7 if(envelopes.length==1) 8 return 1; 9 A[] array = new A[envelopes.length]; 10 for(int i=0;i<envelopes.length;i++){ 11 array[i] =new A(envelopes[i][0],envelopes[i][1]); 12 } 13 //为所有信封按“大小”排序 14 MyCompatator m = new MyCompatator(); 15 Arrays.sort(array,m); 16 int[] result = new int[envelopes.length]; 17 18 int max =1; 19 //很简单的循环操作找最大值,不在细讲 20 for(int i=0;i<envelopes.length;i++){ 21 result[i]=1; 22 for(int j=0;j<i;j++){ 23 if( (array[j].w<array[i].w) && (array[j].h<array[i].h) ){ 24 result[i] = Math.max(result[i], 1+result[j]); 25 } 26 } 27 if(max<result[i]){ 28 max=result[i]; 29 } 30 } 31 return max; 32 } 33 } 34 //比较器,用来为数组排序 35 class MyCompatator implements Comparator<A>{ 36 37 @Override 38 public int compare(A o1, A o2) { 39 if(o1.w>o2.w){ 40 return 1; 41 } 42 else if(o1.w<o2.w){ 43 return -1; 44 } 45 else{ 46 if(o1.h>o2.h){ 47 return 1; 48 } 49 else if(o1.h<o2.h){ 50 return -1; 51 } 52 else 53 return 0; 54 } 55 } 56 57 } 58 //信封类 59 class A{ 60 int w; 61 int h; 62 public A(int w,int h){ 63 this.w=w; 64 this.h=h; 65 } 66 }
时间: 2024-11-07 20:52:21