leetcode Median of Two Sorted Arrays

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)).

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

题意：给定两个已排序数组，找出中间的数（即可以将两个有序数组分成左右数目相同的两部分的数）

思路：使用二分搜索从较短的数组中找出合适的划分，时间复杂度为O(min(len1, len2))。

class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int len1 = nums1.size(), len2 = nums2.size();
if (len1 < len2) return findMedianSortedArrays(nums2, nums1);

int low = 0, high = len2 * 2;//从较短数组的最大元素开始
while (low <= high) {
int mid2 = (low + high) / 2;
int mid1 = len1 + len2 - mid2;//确保划分后左侧元素个数等于右侧元素个数
//第一次划分，分别取较长数组和较短数组的中间数
double l1 = (mid1 == 0) ? INT_MIN : nums1[(mid1-1)/2];
double l2 = (mid2 == 0) ? INT_MIN : nums2[(mid2-1)/2];
double r1 = (mid1 == len1 * 2) ? INT_MAX : nums1[mid1/2];
double r2 = (mid2 == len2 * 2) ? INT_MAX : nums2[mid2/2];

if (l1 > r2) low = mid2 + 1;//若较长数组的左侧最大值大于较短数组的右侧最小值，说明较短数组划分位置应右移，即low=mid2+1；
else if (l2 > r1) high = mid2 - 1;//若较短数组的左侧最大值大于较长数组的右侧最小值，说明较短数组的划分位置应左移，即high=mid2-1；
else return (max(l1,l2) + min(r1, r2)) / 2;
}
return -1;
}
};