Median of Two Sorted Arrays

There are two sorted arrays A and B 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)).

O(m + n) 解法：

 1 public class Solution {
 2     public double findMedianSortedArrays(int A[], int B[]) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         double result = 0;
 6         int lengthA = A.length;
 7         int lengthB = B.length;
 8         
 9         int[] combinedArray = new int[lengthA + lengthB];
10         int i = 0;
11         int j = 0;
12         while(i < lengthA && j < lengthB){
13             if(A[i] > B[j]){
14                 combinedArray[i + j] = B[j];
15                 j ++;
16             }else{
17                 combinedArray[i + j] = A[i];
18                 i ++;
19             }
20         }
21         if(i == lengthA){
22             while(j < lengthB)
23                 combinedArray[i + j] = B[j ++];
24         }
25         if(j == lengthB){
26             while(i < lengthA)
27                 combinedArray[i + j] = A[i ++];
28         }
29         
30         if(((lengthA + lengthB) % 2) !=0){
31             result = combinedArray[(lengthA + lengthB) / 2];
32             //return combinedArray[(lengthA + lengthB) / 2];
33         } else {
34             result = (combinedArray[(lengthA + lengthB) / 2] + combinedArray[(lengthA + lengthB) / 2 - 1]) / 2.0;
35             //return (combinedArray[(lengthA + lengthB) / 2] + combinedArray[(lengthA + lengthB) / 2 - 1]) / 2;
36         }
37         return result;
38     }
39 }

O(log (m+n)) 解法，肯定是二分：网上解法如下

 1 public class Solution {
 2     public double findMedianSortedArrays(int A[], int B[]) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int aLen = A.length;
 6         int bLen = B.length;
 7         if((aLen + bLen) % 2 ==0){
 8             return (getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2) + getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2 + 1)) / 2.0;
 9         } else {
10             return getKthElement(A, 0, aLen - 1, B, 0, bLen - 1, (aLen + bLen) / 2 + 1);
11         }
12     }
13     
14     public int getKthElement(int A[], int aBeg, int aEnd, int B[], int bBeg, int bEnd, int k){
15         if(aBeg > aEnd){
16             return B[bBeg + (k - 1)];
17         }
18         if(bBeg > bEnd){
19             return A[aBeg + (k - 1)];
20         }
21         
22         int aMid = (aBeg + aEnd) >> 1;
23         int bMid = (bBeg + bEnd) >> 1;
24         int len = aMid - aBeg + bMid - bBeg + 2;
25         
26         if(len > k){
27             if(A[aMid] < B[bMid]){
28                 return getKthElement(A, aBeg, aEnd, B, bBeg, bMid - 1, k);
29             } else {
30                 return getKthElement(A, aBeg, aMid - 1, B, bBeg, bEnd, k);
31             }
32         } else {
33             if(A[aMid] < B[bMid]){
34                 return getKthElement(A, aMid + 1, aEnd, B, bBeg, bEnd, k - (aMid - aBeg + 1));
35             } else {
36                 return getKthElement(A, aBeg, aEnd, B, bMid + 1, bEnd, k - (bMid - bBeg + 1));
37             }
38         }
39     } 
40 }

 第二遍：

 

 1 public class Solution {
 2     public double findMedianSortedArrays(int A[], int B[]) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         int alen = A.length;
 6         int blen = B.length;
 7         if((alen + blen) % 2 == 1){
 8           return findKth(A, B, 0, alen - 1, 0, blen - 1, (alen + blen + 1) / 2);
 9         }else{//当长度为偶数时，midian 是中间两个数之和除二。
10           return (findKth(A, B, 0, alen - 1, 0, blen - 1, (alen + blen) / 2) + findKth(A, B, 0, alen - 1, 0, blen - 1, (alen + blen + 2) / 2)) / 2.0;
11         }
12     }
13     
14     public int findKth(int a[], int b[], int as, int ae, int bs, int be, int n){
15         if(as > ae) return b[bs + n - 1];
16         if(bs > be) return a[as + n - 1];
17         int amid = (as + ae) / 2;
18         int bmid = (bs + be) / 2;
19         int len = amid - as + bmid - bs + 2;// 不能用全长 / 2， 因为前一半并不一定是整数。
20         if(a[amid] > b[bmid]){
21             if(n >= len){// 必须有等号， 因为a[amid] > b[bmid]， 所以b[bmid] 一定不是要取的值。
22                 return findKth(a, b, as, ae, bmid + 1, be, n - (bmid - bs + 1));//每次只能去掉两个数组中某一个的一半。
23             }
24             else{
25                 return findKth(a, b, as, amid - 1, bs, be, n);
26             }
27         }else{
28             if(n >= len){
29                 return findKth(a, b, amid + 1, ae, bs, be, n - (amid - as + 1));
30             }
31             else{
32                 return findKth(a, b, as, ae, bs, bmid - 1, n);
33             }
34         }
35     }
36 }

 

posted on 2013-03-26 11:48  Step-BY-Step  阅读(147)  评论(0)  编辑  收藏  举报

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