[leetcode] 4. Median of Two Sorted Arrays

题目链接: https://leetcode.com/problems/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)).

思路

经典的题目

原理和一个数组中求中位数的二分搜索相似。要找到这样一个十字交叉的位置, 也就是i+j = n/2, 使得A[i-1] <= B[j], B[j-1] <= A[i], 这样对于只有一个中位数的取max(A[i-1], B[j-1]), 对于两个中位数的则取(max(A[i-1], B[j-1]), min(A[i], B[j]))/2;

    class Solution {      public:          double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {              if(nums1.size() > nums2.size()) nums1.swap(nums2);              int len1=nums1.size(), len2=nums2.size();              int mid=(len1+len2-1)/2, left=0, right=len1-1;              while(left <= right)              {                  int mid1 = (left+right)/2, mid2 = mid - mid1;                  if(nums1[mid1] >= nums2[mid2]) right = mid1-1;                  else left = mid1+1;              }              double med1=max(left>0?nums1[right]:INT_MIN, nums2[mid-left]);              if((len1+len2)%2) return med1;              double med2=min(left<len1?nums1[left]:INT_MAX,                               mid-right<len2?nums2[mid-right]:INT_MAX);              return (med1 + med2)/2;          }      };