leetcode 练习题 -- 4. Median of Two Sorted Arrays

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本文介绍了一个算法问题：如何找到两个已排序数组的中位数，并确保整体运行时间复杂度为O(log(m+n))。通过示例说明了不同情况下的中位数计算方法。

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Description

There are two sorted arrays nums1 andnums2 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)).

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

class Solution {
public:
	double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
		int size = nums1.size() + nums2.size();
		int flag = size % 2 == 0 ? 0 : 1;
		double a, b;
		int i = 0, j = 0, sum = 1;
		while (i<nums1.size() && j<nums2.size())
		{
			if (nums1[i] <= nums2[j])
			{
				if (sum == size / 2)
					a = nums1[i];
				if (sum == size / 2 + 1)
				{
					b = nums1[i];
					if (flag)return b;
					else return (a + b) / 2;
				}
				i++; sum++;
			}
			else
			{
				if (sum == size / 2)
					a = nums2[j];
				if (sum == size / 2 + 1)
				{
					b = nums2[j];
					if (flag)return b;
					else return (a + b) / 2;
				}
				j++; sum++;
			}
		}
		while (i<nums1.size())
		{
			if (sum == size / 2)
				a = nums1[i];
			if (sum == size / 2 + 1)
			{
				b = nums1[i];
				if (flag)return b;
				else return (a + b) / 2;
			}
			i++; sum++;
		}
		while (j<nums2.size())
		{
			if (sum == size / 2)
				a = nums2[j];
			if (sum == size / 2 + 1)
			{
				b = nums2[j];
				if (flag)return b;
				else return (a + b) / 2;
			}
			j++; sum++;
		}
	}
};