[leetcode] 4. 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

非常有趣的二分查找题目，首先假设，这个中位数一定来自数组nums1，那么有如下：

1. 在nums1里面二分下标，对每个下标对应的元素，计算其在nums2中最靠前能排到多少名，最靠后能排到多少名，主要是靠重复元素的存在。那么此元素在nums1里面的排名加上其在nums2里面的排名区间，即可判断此元素是否就是中位数。

2. 还需要假设此元素在nums2中

3. 需要考虑合并后的长度是奇数还是偶数

#include<bits/stdc++.h>
using namespace std;

double bsearch(vector<int>& nums1, vector<int>& nums2, int pos) {
    int low = 0,high = nums1.size()-1, tar = -1;
    while(low <= high) {
        int mid = (low + high) >> 1;
        int val = nums1[mid];
        int x = lower_bound(nums2.begin(),nums2.end(),val) - nums2.begin();
        int y = upper_bound(nums2.begin(), nums2.end(),val) - nums2.begin();
        if (mid + x <= pos and pos <= mid + y) {
            tar = nums1[mid];
            break;
        } else if (mid + x > pos) high = mid - 1;
        else low = mid + 1;
    }
    return tar;
}


double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    int tlen = nums1.size() + nums2.size();
    double ret = 0;

    int pos = tlen>>1;
    ret = bsearch(nums1, nums2, pos);
    if (ret == -1) ret = bsearch(nums2, nums1, pos);

    if ((tlen&1) == 0) {
        pos = (tlen>>1)-1;
        double tmp = bsearch(nums1, nums2, pos);
        if (tmp == -1) tmp = bsearch(nums2, nums1, pos);
        ret += tmp;
        ret *= 0.5;
    }

    return ret;

}



int main() {
    vector<int>a = {1,2};
    vector<int>b = {3,4};
    cout<<findMedianSortedArrays(a,b)<<endl;
    return 0;
}