Leetcode 004 Median of Two Sorted Arrays

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Leetcode 004 Median of Two Sorted Arrays

写出给定的两个数组中所有数字的中位数

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题目

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

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My Solution (O(n) not O(logn))

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size(), len2 = nums2.size();
        // if (len1 == 0 && len2 == 0) return 0;
        int position = (len1 + len2) / 2;
        bool odd = ((len1 + len2) % 2 == 1);
        // so postion is 
        // odd=false: (position-1, postion)/2
        // odd=true: position
        int curr = 0;
        int lst, now;
        int i = 0, j = 0;
        while (i < len1 || j < len2) {
            lst = now;
            if (i >= len1) {
                now = nums2[j];
                j++; 
            } else if (j >= len2) {
                now = nums1[i];
                i++; 
            } else {
                if (nums1[i]<nums2[j]) {now = nums1[i]; i++;}
                else {now = nums2[j]; j++;}             
            }           
            if (curr == position) {
                double re = odd ? (double)now : (double)(lst + now) / 2;
                return re;
            }
            curr++;
        }
    }
};

分析

思路很简单，其实就是把两个数组归并为一个数组，然后根据得到这个数组的中位数。

在实际中进行了一些优化：首先，不需要申请空间存储这个归并后的数组，只需要得到用来计算中位数的1或2个数字即可（即数字总数为奇数和偶数时）；其次，不需要将数组的所有都计算出来，只需计算到中间位置即可。

因此该方法并没有达到真正的$O(log(m+n))$，而是要遍历$(m+n)/2$个元素，即$O((m+n)/2)$。