LeetCode 4. Median of Two Sorted Arrays

LeetCode 4. Median of Two Sorted Arrays

Description:

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)).
Difficulty: Hard

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分析如下：

题目要求时间复杂度为O(log(m+n))，应该是这道题难度为Hard的原因。然而，我的解法，也是比较常见的解法，先把两个数组合并成一个，然后排序，找出中位数即可。排序复杂度为O(log(m+n)，合并复杂度为O(m)和O(n)，提交上去还是可以通过的。
看了一下LeetCode的Solution，把数学运用得出神入化，啧啧！
LeetCode Solution
膜！

代码如下：

My code：

class Solution {
public:
	double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
		vector<int> nums;
		for (int i = 0; i < nums1.size(); i++) {
			nums.push_back(nums1[i]);
		}
		for (int i = 0; i < nums2.size(); i++) {
			nums.push_back(nums2[i]);
		}
		sort(nums.begin(), nums.end());
		int len = nums.size();
		double res;
		if (len % 2 == 0) {
			res = double(nums[len / 2 - 1] + nums[len / 2]) / 2;
		}
		else {
			res = double(nums[len / 2]);
		}
		return res;
	}
};

LeetCode Solution

// Java
class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) { // to ensure m<=n
            int[] temp = A; A = B; B = temp;
            int tmp = m; m = n; n = tmp;
        }
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j-1] > A[i]){
                iMin = iMin + 1; // i is too small
            }
            else if (i > iMin && A[i-1] > B[j]) {
                iMax = iMax - 1; // i is too big
            }
            else { // i is perfect
                int maxLeft = 0;
                if (i == 0) { maxLeft = B[j-1]; }
                else if (j == 0) { maxLeft = A[i-1]; }
                else { maxLeft = Math.max(A[i-1], B[j-1]); }
                if ( (m + n) % 2 == 1 ) { return maxLeft; }

                int minRight = 0;
                if (i == m) { minRight = B[j]; }
                else if (j == n) { minRight = A[i]; }
                else { minRight = Math.min(B[j], A[i]); }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
}

# python
def median(A, B):
    m, n = len(A), len(B)
    if m > n:
        A, B, m, n = B, A, n, m
    if n == 0:
        raise ValueError

    imin, imax, half_len = 0, m, (m + n + 1) / 2
    while imin <= imax:
        i = (imin + imax) / 2
        j = half_len - i
        if i < m and B[j-1] > A[i]:
            # i is too small, must increase it
            imin = i + 1
        elif i > 0 and A[i-1] > B[j]:
            # i is too big, must decrease it
            imax = i - 1
        else:
            # i is perfect

            if i == 0: max_of_left = B[j-1]
            elif j == 0: max_of_left = A[i-1]
            else: max_of_left = max(A[i-1], B[j-1])

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

            if i == m: min_of_right = B[j]
            elif j == n: min_of_right = A[i]
            else: min_of_right = min(A[i], B[j])

            return (max_of_left + min_of_right) / 2.0