C++博客-程序描绘人生-随笔分类-C++

LeetCode – Median of Two Sorted Arrays - findMedianSortedArrays

胡满超 — Tue, 26 Jun 2018 05:57:00 GMT

There are two sorted arrays A and B 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)).

The following code is better than most of the results returned by baidu or google. Time complexity is O((m+n)/2), Space complexity is O(1).

1 double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
2 {
3     int nums1_i = 0, nums2_i = 0;
4     int mid1 = 0, mid2 = 0, count = 0;
5     while (nums1_i < nums1.size() && nums2_i < nums2.size())
6     {
7         if (count++ > ((nums1.size() + nums2.size()) / 2))    break;
8         mid1 = mid2;
9         mid2 = (nums1[nums1_i] < nums2[nums2_i] ? nums1[nums1_i++] : nums2[nums2_i++]);
10     }
11
12     while (nums1_i < nums1.size())
13     {
14         if (count++ > ((nums1.size() + nums2.size()) / 2))    break;
15         mid1 = mid2;
16         mid2 = nums1[nums1_i++];
17     }
18
19     while (nums2_i < nums2.size())
20     {
21         if (count++ > ((nums1.size() + nums2.size()) / 2))    break;
22         mid1 = mid2;
23         mid2 = nums2[nums2_i++];
24     }
25
26     return (nums1.size() + nums2.size()) % 2 == 0
27         ? (mid1 + mid2) / 2.0
28         : mid2;
29 }

胡满超 2018-06-26 13:57 发表评论

solution to the complier error "cannot open file 'libboost_thread-vcxx-mt-sgd-xx.lib'"

胡满超 — Tue, 20 Aug 2013 07:52:00 GMT

You need compile boost and add libs path in project or makefile.

Compile boost run:

bjam stage --toolset=msvc-10.0 link=static runtime-link=static threading=multi debug release

胡满超 2013-08-20 15:52 发表评论

gdb 查看死锁

胡满超 — Tue, 20 Aug 2013 07:40:00 GMT

转自：http://blog.chinaunix.net/uid-23766031-id-2386460.html

死锁：一种情形，此时执行程序中两个或多个线程发生永久堵塞（等待），每个线程都在等待被

其他线程占用并堵塞了的资源。例如，如果线程A锁住了记录1并等待记录2，而线程B锁住了记录2并等待记录1，这样两个线程就发生了死锁现象。

gdb调试死锁的方法：

gdb

attach pid

thread apply all bt

找到_lll_lock_wait 锁等待的地方。

然后查找该锁被哪个线程锁住了。

例如：

查看哪个线程拥有互斥体（然后list代码，查看使用互斥变量的名称）

(gdb) print AccountA_mutex

$1 = {__m_reserved = 2, __m_count = 0, __m_owner = 0x2527,

__m_kind = 0, __m_lock

= {__status = 1, __spinlock = 0}}

(gdb) print 0x2527

$2 = 9511

(gdb) print AccountB_mutex

$3 = {__m_reserved = 2, __m_count = 0, __m_owner = 0x2529,

__m_kind = 0, __m_lock = {__status = 1, __spinlock = 0}}

(gdb) print 0x2529

$4 = 9513

(gdb)

从上面的命令中，我们可以看出AccontA_mutex是被线程 5（LWP 9511）加锁（拥有）的，而AccontB_mutex是被线程 3（LWP 9513）加锁（拥有）的。

胡满超 2013-08-20 15:40 发表评论