<p>Given two sorted arrays <code>nums1</code> and <code>nums2</code> of size <code>m</code> and <code>n</code> respectively, return <strong>the median</strong> of the two sorted arrays.</p>

<p>The overall run time complexity should be <code>O(log (m+n))</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,3], nums2 = [2]
<strong>Output:</strong> 2.00000
<strong>Explanation:</strong> merged array = [1,2,3] and median is 2.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums1 = [1,2], nums2 = [3,4]
<strong>Output:</strong> 2.50000
<strong>Explanation:</strong> merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>nums1.length == m</code></li>
	<li><code>nums2.length == n</code></li>
	<li><code>0 &lt;= m &lt;= 1000</code></li>
	<li><code>0 &lt;= n &lt;= 1000</code></li>
	<li><code>1 &lt;= m + n &lt;= 2000</code></li>
	<li><code>-10<sup>6</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size(), n2 = nums2.size();
        if(n1>n2) return findMedianSortedArrays(nums2, nums1);
        int low = 0, high = n1;
        int left = (n1+n2+1)/2;
        int n = n1+n2;
        while(low <= high)
        {
            int mid1 = low+ (high-low)/2;
            int mid2 = left-mid1;
            int l1 = INT_MIN, l2 = INT_MIN;
            int r1 = INT_MAX, r2 = INT_MAX;
            if(mid1 < n1) r1 = nums1[mid1];
            if(mid2 < n2) r2 = nums2[mid2];
            if(mid1 - 1 >= 0) l1 = nums1[mid1-1];
            if(mid2 - 1 >= 0) l2 = nums2[mid2-1];
            if(l1 <= r2 && l2 <=r1)
            {
                if(n%2 == 1) return max(l1,l2);
                else return (double)(max(l1,l2) + min(r1,r2))/2.0;
            }
            else if(l1 > r2) high = mid1-1;
            else low = mid1 + 1;
        }
        return 0.0;
    }
};
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