[leetcode]4. Median of Two Sorted Arrays俩有序数组的中位数

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)).

You may assume nums1 and nums2 cannot be both empty.

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

Solution1: time complexity is O(m+n). merge two sorted array, then find the median

code:

 /*
   Time Complexity:  O(m+n)
   Space Complexity: O(m+n)
 */
 
 class Solution {
     public double findMedianSortedArrays(int[] nums1, int[] nums2) {   // find median
         int[] mergedArray = mergeTwoSortedArray(nums1, nums2);
         int n = mergedArray.length;
         if( n % 2 == 0){
             return (mergedArray[(n-1)/2] + mergedArray[n/2])/2.0;
         }else{
             return mergedArray[n/2];
         }
     } 
 
     public int[] mergeTwoSortedArray(int[] nums1, int[] nums2){ // merge sort
         int[] merged = new int[nums1.length + nums2.length];
         int i = 0, j = 0, k = 0;
         // 两个array同时扫
         while( i < nums1.length && j < nums2.length){
             merged[k++] = (nums1[i] < nums2[j]) ? nums1[i++] : nums2[j++];
         }
         //扫到只剩下较长的那个array， either nums1 or nums2
         while ( i < nums1.length ){
              merged[k++] = nums1[i++];
         }
         while(j < nums2.length){
             merged[k++] = nums2[j++];
         }
         return merged;
     }
 }

Solution2: time complexity is O(log (m+n)).

这个题的思路可以是找出2个sorted array 所有元素中，Kth 元素

因此我们可以把问题转化成，

“Given 2 sorted arrays, A, B of sizes m, n respectively, find the numbers which are NOT medians of the two sorted arrays”

如果我们能找到那些NOT medians的数字，删掉，并不断缩小范围，那最终剩下的一定就是actual median

Step1, 确定要找的median是merged之后array中的第几个元素

Step2, 用binary search切开nums1, nums2。发现nums1左半边的长度为4， nums2左半边的长度为2。

Step3, nums1左半边长度+ nums2左半边长度为6

Step4， 所以第7个元素可能在nums1的右半边pivot上，或者在nums2的右半边pivot上

比较发现，第7个元素应该是7

code：

 /*
  Time Complexity: O(log(m+n))
  Space Complexity: O(log(m+n))
 */
 
 class Solution {
     public double findMedianSortedArrays(final int[] A, final int[] B) {
         int total = A.length + B.length;
         if (total %2 == 0){
             return (findKth(A, 0, B, 0, total / 2) + findKth(A, 0, B, 0, total / 2 + 1)) / 2.0;
         }else{
             return findKth(A, 0, B, 0, total / 2 + 1);
         }
     }
 
     private static int findKth(final int[] A, int ai, final int[] B, int bi, int k) {
         //always assume that A is shorter than B
         if (A.length - ai > B.length - bi) {
             return findKth(B, bi, A, ai, k);
         }
         if (A.length - ai == 0) return B[bi + k - 1];
         if (k == 1) return Math.min(A[ai], B[bi]);
 
         //divide k into two parts
         int k1 = Math.min(k / 2, A.length - ai), k2 = k - k1;
         if (A[ai + k1 - 1] < B[bi + k2 - 1])
             return findKth(A, ai + k1, B, bi, k - k1);
         else if (A[ai + k1 - 1] > B[bi + k2 - 1])
             return findKth(A, ai, B, bi + k2, k - k2);
         else
             return A[ai + k1 - 1];
     }
 }