LeetCode——004-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)).

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
当一看到这道题的算法复杂度的限制为O(log(m+1)),
就可以排除将两个数组排序后再寻找中位数。
感觉题目又在提示我们使用二分法，但感觉此处没法使用二分法。

我的方法就是一边排序一遍判断是否是位，即使用k来对数组A，B进行排序，
当k==（A+B）/2【数组大小】时，即可直接返回中位数，唯一值得注意的就是数组大小为奇数还是偶数的问题。

class Solution {
public:
	double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
		int m = nums1.size(), n = nums2.size();
		int i = 0, j = 0, k = 0, mid = (m + n) / 2 + ((m + n) % 2 == 0 ? 0 : 1);
		double sum = 0.0;
		while (i < m && j < n)
		{
			if (nums1[i] <= nums2[j])sum += nums1[i++];
			else sum += nums2[j++];
			++k;
			if (isReturn(sum, k, mid, m + n))return sum;
		}
		while (i < m)
		{
			sum += nums1[i++];
			++k;
			if (isReturn(sum, k, mid, m + n))return sum;
		}
		while (j < n)
		{
			sum += nums2[j++];
			++k;
			if (isReturn(sum, k, mid, m + n))return sum;
		}
		return sum;
	}
	bool isReturn(double &sum, int k, int mid, int s)
	{
		if (k < mid)
		{
			sum = 0.0;
			return false;
		}
		else if (k == mid && s % 2 == 1)
			return true;
		else if (k == mid + 1)
		{
			sum /= 2.0;
			return true;
		}
		return false;
	}
};

借鉴
借鉴一下大神的思路,
使用递归求解，但大致思路和我一样。

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size(), left = (m + n + 1) / 2, right = (m + n + 2) / 2;
        return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
    }
    int findKth(vector<int>& nums1, int i, vector<int>& nums2, int j, int k) {
        if (i >= nums1.size()) return nums2[j + k - 1];
        if (j >= nums2.size()) return nums1[i + k - 1];
        if (k == 1) return min(nums1[i], nums2[j]);
        int midVal1 = (i + k / 2 - 1 < nums1.size()) ? nums1[i + k / 2 - 1] : INT_MAX;
        int midVal2 = (j + k / 2 - 1 < nums2.size()) ? nums2[j + k / 2 - 1] : INT_MAX;
        if (midVal1 < midVal2) {
            return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
        } else {
            return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
        }
    }
};