Median of Two Sorted Arrays
来源:互联网 发布:关联规则算法图 编辑:程序博客网 时间:2024/06/05 06:15
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
Example 1:
nums1 = [1, 3]nums2 = [2]The median is 2.0
Example 2:
nums1 = [1, 2]nums2 = [3, 4]The median is (2 + 3)/2 = 2.5
package November;public class MedianofTwoSortedArrays {public static double findMedianSortedArrays(int[] nums1, int[] nums2) {/** * 假设我们可以采用类似于merge的方式,用两个指针分别指向数组A和B,两个指针依次比较,这种时间复杂度是O(n) * 需要找到O(logN)时间复杂度的算法,题目思路是将寻找中位数转换为,寻找两个有序数组的第k小的数 * 如何求第k小的数呢?A[k/2]和B[k-k/2]进行比较, * 如果A[k/2]>B[k-k/2],则B[0]到B[k-k/2]范围内的数据都可以被抛弃掉,A[k/2+1]到A[tail]之间的数据都可以抛弃掉, * 问题转换为寻找A[0]-A[k/2]和B[k-k/2+1]-B[tail]中第k-(k-k/2+1)小的数 * 如果A[k/2]<B[k-k/2],情况类似 * 如果A[k/2]==B[k-k/2],则第k小的数已经找到 * 对于m+n为奇数的情况,我们只需要寻找(m+n)/2小的数,并返回 * 对于m+n为偶数的情况,我们需要寻找(m+n)/2和(m+n)/2-1小的数,返回平均值 *///注意最开始就应该有顺序int start1=0;int end1=nums1.length-1;int start2=0;int end2=nums2.length-1;int middle = (nums1.length+nums2.length)/2;if((nums1.length+nums2.length)%2==1){return findMedianSortedArrays(nums1,start1,end1,nums2,start2,end2,middle+1);}else{return (findMedianSortedArrays(nums1,start1,end1,nums2,start2,end2,middle)+findMedianSortedArrays(nums1,start1,end1,nums2,start2,end2,middle+1))/2;}}public static double findMedianSortedArrays(int[] nums1,int start1,int end1,int[] nums2, int start2,int end2,int k){// 假设nums1数组是那个比较短的数组System.out.println(start1+","+end1+","+start2+","+end2+","+k);if(end1-start1>end2-start2){// 这个if应该在最上面return findMedianSortedArrays(nums2,start2,end2,nums1,start1,end1,k);}if(start1>end1){ // if条件的顺序,这个必须在k==1之前return nums2[start2+k-1];}if(k==1){return Math.min(nums1[start1], nums2[start2]); }int pa=Math.min(k/2,end1-start1+1);int pb=k-pa;if(nums1[start1+pa-1]>nums2[start2+pb-1]){return findMedianSortedArrays(nums1,start1,start1+pa-1,nums2,start2+pb,end2,k-pb);}else if(nums1[start1+pa-1]<nums2[start2+pb-1]){return findMedianSortedArrays(nums1,start1+pa,end1,nums2,start2,start2+pb-1,k-pa);}else{return nums1[start1+pa-1];}}public static void main(String[] args){//int[] arr1={5,6,9};//int[] arr2={1,1,2,3,4};//int[] arr2={5,6,9};//int[] arr1={1,1,2,3,4};//int[] arr2={0,5,9,10,14};//int[] arr1={1};//int[] arr1={1,2};//int[] arr2={3,4};//int[] arr1={1};//int[] arr2={};int[] arr1={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22};int[] arr2={0,6};System.out.println(findMedianSortedArrays(arr1,arr2));}}
阅读全文
0 0
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of two sorted arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- median-of-two-sorted-arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Median of Two Sorted Arrays
- Suddenly Bianca's face cleared, and she looked at me with steady determination.
- debian下安装LNMP环境(一)
- MongoDB:使用keyfile访问控制的方式部署副本集
- 字符串拆分的一个小例子
- [Kafka]
- Median of Two Sorted Arrays
- postgresql query user
- js中this指向问题
- 简易——CentOS配置单个tomcat到linux服务器上
- Hibernate通常是三种:hql查询,QBC查询和QBE查询:
- 学术 | DeepMind最新研究:使用更简单的环境就能检测AI是否安全
- CBinsight重磅报告 | 如何从谷歌亚马逊苹果微软脸书的9年专利之争,看5大巨头在AI行业的未来10年之争
- 观点 | Keras之父谈人工智能:通用AI不会出现,超人类智能更不存在
- BDTC 2017 | 中国大数据技术大会全日程和讲师曝光