Median of Two Sorted Arrays-分治法

题目意思很简单将两个有序数组合并之后的中位数找出来。题目要求使用log(m+n)的时间复杂度来做。

虽然言简意赅,但不得不承认这个题目我自己想了好久也没做出来,隐约觉得应该使用寻找第k大数的算法来做,但是具体到这个题目,编码多次都以失败告终,所以不得不去网上参考下别人的思路和代码。

参考链接:http://blog.csdn.net/zxzxy1988/article/details/8587244

但是这个方法现在已经无法在leetcode上AC,需要在两个递归的return处进行修改。

思路是这样的:

我们寻找两个数组的第k个数,那么我们首先找到两个数组中的第(k/2-1)个数。比较两个数组中这个数的大小来进行不同递归。

代码如下:

 1     int min(int a,int b)
 2     {
 3         return a>b?b:a;
 4     }
 5     double findknumber(int *a,int *b,int al,int bl,int k)
 6     {
 7         if (al > bl)
 8             return findknumber(b, a, bl, al, k);
 9         if (al == 0)
10             return b[k - 1];
11         if (k == 1)
12             return min(a[0], b[0]);
13         int aindex = min(k / 2, al);
14         int bindex = k - aindex;
15         if (a[aindex - 1] < b[bindex - 1])
16             return findknumber(a + aindex, b, al - aindex, bindex, k - aindex);
17         else if (a[aindex - 1] > b[bindex - 1])
18             return findknumber(a, b + bindex, aindex, bl - bindex, k - bindex);
19         else
20             return a[aindex - 1];
21
22
23     }
24     double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
25         int total=nums1Size+nums2Size;
26         if(total%2==1)
27             return findknumber(nums1,nums2,nums1Size,nums2Size,total/2+1);
28         else
29             return (findknumber(nums1,nums2,nums1Size,nums2Size,total/2+1)
30                     +findknumber(nums1,nums2,nums1Size,nums2Size,total/2))/2.0;
31
32     }

这个代码还存在两个问题,后续我再补充:

存在一个找第k个数的坐标问题;

这个算法的真实时间复杂度如何;

时间: 2024-10-13 02:15:34

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