Median of Two Sorted Arrays——解题笔记

客官°小女子只卖身不卖艺 2022-08-08 05:26 137阅读 0赞

【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; 
    } 
    };