1. Question
找两个有序数组的中位数。时间复杂度O(log(m+n))
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)).
2. Solution
2.1 O(nlogn)
将两个数组拷到新数组中，对新数组排序
1 public class Solution { 2 public double findMedianSortedArrays( int[] a, int[] b ){ 3 int[] c = new int[a.length + b.length]; 4 System.arraycopy(a, 0, c, 0, a.length); 5 System.arraycopy(b, 0, c, a.length, b.length); 6 Arrays.sort(c); 7 int mid = c.length / 2; 8 if( c.length % 2 == 0 ) 9 return ( c[mid-1] + c[mid] ) / 2.0; 10 return c[mid]; 11 } 12 }
2.2 O(n/2)---->O(n)
将两个数组合并。
1 public class Solution { 2 public double findMedianSortedArrays( int[] a, int[] b ){ 3 int len1 = a.length; 4 int len2 = b.length; 5 int len = (len1 + len2) / 2 + 1; 6 int[] c = new int[len]; 7 int ia; 8 int ib; 9 int ic; 10 for( ia=0, ib=0, ic=0; ia<len1 && ib<len2 && ic<len; ic++ ){ 11 if( a[ia] <= b[ib] ) 12 c[ic] = a[ia++]; 13 else 14 c[ic] = b[ib++]; 15 } 16 for( ; ia<len1 && ic<len; ic++, ia++ ) 17 c[ic] = a[ia]; 18 for( ; ib<len2 && ic<len; ic++, ib++ ) 19 c[ic] = b[ib]; 20 ic--; 21 if( (len1 + len2) % 2 == 0 ) 22 return ( c[ic] + c[ic-1] ) / 2.0; 23 return c[ic]; 24 } 25 }
2.3 O(logn)
定义如下变量：
- m1：nums1的中位数（有m/2个数小于等于m1）
- m2：nums2的中位数（有n/2个数小于等于m2）
判断m1和m2的大小，以舍弃一半的数组：
- m1<m2：有(m+n)/2个数<=m2，即m1必然存在于合并后数组的前半部分，m2存在于后半部分。则中位数存在于nums1的后半部分或nums2的前半部分；
- m1>m2：同上，中位数存在于nums1的前半部分或nums2的后半部分；
- m1 == m2：m1即是中位数
1 public class Solution { 2 // find the kth big number; 3 public int findKth( int[] a, int afrom, int ato, int[] b, int bfrom, int bto, int k ){ 4 //we assume that m<=n 5 int m = ato - afrom + 1; 6 int n = bto - bfrom + 1; 7 if( m > n ) 8 return findKth( b, bfrom, bto, a, afrom, ato, k); 9 if( m == 0 ) 10 return b[k-1]; 11 if( k==1 ) 12 return ( a[afrom] < b[bfrom] ) ? a[afrom] : b[bfrom]; 13 int na = ( k/2 < m )? k/2 : m; 14 int nb = k-na; 15 int pa = na + afrom -1; 16 int pb = nb + bfrom -1; 17 if( a[pa] > b[pb] ) 18 return findKth( a, afrom, pa, b, pb+1, bto, k-nb ); 19 if( a[pa] < b[pb] ) 20 return findKth( a, pa+1, ato, b, bfrom, pb, k-na ); 21 return b[pb]; 22 } 23 24 //O( log(m+n) ) time 25 public double findMedianSortedArrays( int[] a, int[] b ){ 26 int len = a.length + b.length; 27 if( len % 2 == 0 ) 28 return ( findKth( a, 0, a.length-1, b, 0, b.length-1, len/2 ) + findKth( a, 0, a.length-1, b, 0, b.length-1, len/2+1 ) ) / 2.0; 29 else 30 return findKth( a, 0, a.length-1, b, 0, b.length-1, len/2+1 ); 31 } 32 }