Median of Two Sorted Arrays——解题笔记

【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)).

解法1：

直接把两个vector合并，然后排序，取中位数。注意对合并后vector长度的奇偶数的处理。此解法比较好想到，但是时间复杂度是O((m+n)log(m+n)).

代码如下：

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        vector<int> mergedNums;
        mergedNums.assign(nums1.begin(), nums1.end());
        mergedNums.insert(mergedNums.end(), nums2.begin(), nums2.end());
        sort(mergedNums.begin(), mergedNums.end());

        int len1 = nums1.size();
        int len2 = nums2.size();

        double median = (len1 + len2)%2 ? mergedNums[(len1 + len2)>>1] : (mergedNums[(len1 + len2 - 1)>>1] + mergedNums[(len1 + len2)>>1])/2.0;

        return median;
    }
};

解法2：

时间复杂度能达到O(log(m+n))，但是思路比较复杂。我们可以把这个问题转化为寻找第k小的数，只不过这里的k是中位数。

参考解释：

http://blog.csdn.net/yutianzuijin/article/details/11499917

http://blog.csdn.net/zxzxy1988/article/details/8587244

代码如下：

double findKth(int a[], int m, int b[], int n, int k)
{
//always assume that m is equal or smaller than n
if (m > n)
return findKth(b, n, a, m, k);
if (m == 0)
return b[k - 1];
if (k == 1)
return min(a[0], b[0]);
//divide k into two parts
int pa = min(k / 2, m), pb = k - pa;
if (a[pa - 1] < b[pb - 1])
return findKth(a + pa, m - pa, b, n, k - pa);
else if (a[pa - 1] > b[pb - 1])
return findKth(a, m, b + pb, n - pb, k - pb);
else
return a[pa - 1];
} 

class Solution
{
public:
double findMedianSortedArrays(int A[], int m, int B[], int n)
{
int total = m + n;
if (total & 0x1)
return findKth(A, m, B, n, total / 2 + 1);
else
return (findKth(A, m, B, n, total / 2)
+ findKth(A, m, B, n, total / 2 + 1)) / 2;
}
};