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

Solution: Binary Search

1. 要找median，原始的方法是把两个arrays合并，然后找中位数。但是Time Complexity是O(m + n)
2. 那么O(log (m+n))的要求，就肯定是binary search， 思路如下
   
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class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        
        //1. Binary search: to find partition x &  partition Y, such that
        // 
        // 1) partition x + partition y = (xLen + yLen + 1) / 2, means element number of left half == element number of right half
        // 2) all elements from two arrays left half <= all elements from two arrays right half
        // x1, x2,| x3, x4, x5, x6, 
        // y1, y2, y3, y4, y5,| y6, y7, y8
        // partition x = 2, partition y = 5 ==> left half has 5 elements, right half has 5 elements
        // and make sure x2 <= y6 && y5 <=x3  (coz it's sorted array, therefore already guaranteed x2 <= x3, y5 <= y6)
        
        if (nums1.length > nums2.length) {
            return findMedianSortedArrays (nums2, nums1);
        }
        
        int nums1Len = nums1.length;
        int nums2Len = nums2.length;
        int halfLen = (nums1Len + nums2Len + 1) / 2;
        
        int start = 0;
        int end = nums1Len;
        int partitionX = 0;
        int partitionY = 0;
        
        while (start <= end) {
            partitionX = start + (end - start) / 2;
            partitionY = halfLen - partitionX;
            
            // if partitionX == 0, means there is nothing on the left, use MIN_VALUE instead
            int maxLeftX = partitionX == 0 ? Integer.MIN_VALUE : nums1 [partitionX - 1];
            int maxLeftY = partitionY == 0 ? Integer.MIN_VALUE : nums2 [partitionY - 1];
            
            // if partitionX == nums1 Len, means there is nothing on the right, use MAX_VALUE instead
            int minRightX = partitionX == nums1Len ? Integer.MAX_VALUE : nums1 [partitionX];
            int minRightY = partitionY == nums2Len ? Integer.MAX_VALUE : nums2 [partitionY];
             
            if (maxLeftX <= minRightY && maxLeftY <= minRightX) {
                
                // if combined array len is even
                if ((nums1Len + nums2Len) % 2 == 0) {
                    //System.out.println ("1+++++" + "x: " + partitionX + " Y: " + partitionY);
                    return (Math.max (maxLeftX, maxLeftY) + Math.min (minRightX, minRightY)) / 2.0;
                }

                // if it's odd
                //System.out.println ("2+++++" + "x: " + partitionX + " Y: " + partitionY);
                return (Math.max (maxLeftX, maxLeftY));    
            } else if (maxLeftX > minRightY) {
                end = partitionX - 1;
            } else if (maxLeftY > minRightX) {
                start = partitionX + 1;
            }
        }
        
        return -1;
        
        
    }
}