LeetCode(4),Median of Two Sorted Arrays

Problem:

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:

LeetCode官网已经给出了很好的解答，答案是用java和python写的，这里把它改成php形式。基本思路就是二分查找。

/**
 * @param Integer[] $nums1
 * @param Integer[] $nums2
 * @return Float
 */
function findMedianSortedArrays($nums1, $nums2) {
    $m = count($nums1);
    $n = count($nums2); 
    if($m>$n){
        $temp = $nums2;
        $nums2 = $nums1;
        $nums1 = $temp;
        $t = $n;
        $n = $m;
        $m = $t;
    }

    if($n===0)
        throw new Exception("arrays cann't be empty", 1);

    $imin = 0;$imax = $m; $half_len = (int)(($m+$n+1)/2);
    while($imin<=$imax){
        $i = (int)(($imin+$imax)/2);
        $j = $half_len - $i;
        // echo $i;echo $j;
        if ($i < $m and $nums2[$j-1] > $nums1[$i])
            # i is too small, must increase it
            $imin = $i + 1;
        elseif($i > 0 and $nums1[$i-1] > $nums2[$j])
            # i is too big, must decrease it
            $imax = $i - 1;
        else{
            if($i==0){
                $max_of_left = $nums2[$j-1];
            }elseif($j==0){
                $max_of_left = $nums1[$i-1];
            }else{
                $max_of_left = ($nums1[$i-1]>$nums2[$j-1])?$nums1[$i-1]:$nums2[$j-1];
            }

            if(($m+$n)%2===1)
                return $max_of_left;

            if($m=$i)
                $min_of_right = $nums2[$j];
            elseif($j==$n)
                $min_of_right = $nums1[$i];
            else
                $min_of_right = ($nums1[$i]<$nums2[$j])?$nums1[$i]:$nums2[$j];

            return ($min_of_right + $max_of_left)/2.0; 
        }

    }
          

}