题目

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

Follow up: The overall run time complexity should be O(log (m+n)).

Example 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Example 3:

Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:

Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:

Input: nums1 = [2], nums2 = []
Output: 2.00000

Constraints:

nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106

解法：二分

https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xun-zhao-liang-ge-you-xu-shu-zu-de-zhong-wei-s-114/
将其视作找第K小问题，对K二分！

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int tot = nums1.size() + nums2.size();
        if (tot % 2) 
            return find(nums1, nums2, tot / 2 + 1);
        else {
            int left = find(nums1, nums2, tot / 2);
            int right = find(nums1, nums2, tot / 2 + 1);
            return (left + right) / 2.0;
        }
    }
    double find(vector<int> &nums1, vector<int> &nums2, int k) {
        int m = nums1.size(),  n = nums2.size();
        int i = 0, j = 0;
        while (true) {
            if (i == m) return nums2[j + k - 1];
            if (j == n) return nums1[i + k - 1];
            if (k == 1) return min(nums1[i], nums2[j]);
            int i1 = min(m, i + k / 2), j1 = min(n, j + k / 2);
            if (nums1[i1 - 1] < nums2[j1 - 1]) {
                k -= i1 - i;
                i = i1;
            }
            else {
                k -= j1 - j;
                j = j1;
            }
        }
    }
};