我最初使用蛮力解决了这个challenge并且它被接受了 . 我试图利用memoization动态编程来减少 O(2^n)
的时间复杂度 .
使用memoization的动态编程比蛮力方法花费的时间更长,并且我收到超出时间限制的错误消息 .
蛮力方法代码 .
public class Dummy
{
private int answer = 0;
private int numberCalled = 0;
public bool doFindSum(ref int[] nums, int index, int current, int target)
{
numberCalled++;
if (index + 1 == nums.Length)
{
if (current == target)
{
++answer;
return true;
}
else
{
return false;
}
}
bool add = doFindSum(ref nums, index + 1, current + nums[index + 1], target);
bool minus = doFindSum(ref nums, index + 1, current - nums[index + 1], target);
return add || minus;
}
public int FindTargetSumWays(int[] nums, int S)
{
numberCalled = 0;
doFindSum(ref nums, -1, 0, S);
Console.WriteLine("Nums Called = {0}", numberCalled);
return answer;
}
}
使用记忆代码进行动态编程
public class DP
{
private Dictionary<int, Dictionary<int, int>> dp;
private int numberCalled = 0;
public int doFindSum(ref int[] nums, int index, int current, int target)
{
numberCalled++;
Dictionary<int, int> temp;
if (dp.TryGetValue(index + 1, out temp))
{
int value;
if (temp.TryGetValue(current, out value))
{
return value;
}
}
if (index + 1 == nums.Length)
{
if (current == target)
{
if (!dp.ContainsKey(index + 1))
{
dp.Add(index + 1, new Dictionary<int, int>() { { current, 1 } });
return 1;
}
}
return 0;
}
int add = doFindSum(ref nums, index + 1, current + nums[index + 1], target);
int minus = doFindSum(ref nums, index + 1, current - nums[index + 1], target);
if ((!dp.ContainsKey(index + 1)) && (add + minus) > 0)
{
dp.Add(index + 1, new Dictionary<int, int>() { { current, add + minus } });
}
return add + minus;
}
public int FindTargetSumWays(int[] nums, int S)
{
numberCalled = 0;
dp = new Dictionary<int, Dictionary<int, int>>(); // index , sum - count
var answer = doFindSum(ref nums, -1, 0, S);
Console.WriteLine("Nums Called = {0}", numberCalled);
return answer;
}
}
并且Code为驱动程序编码测量每种方法所花费的时间
public static void Main(string[] args)
{
var ip = new int[][] { new int [] { 0, 0, 0, 0, 0, 0, 0, 0, 1},
new int [] {6,44,30,25,8,26,34,22,10,18,34,8,0,32,13,48,29,41,16,30},
new int []{7,46,36,49,5,34,25,39,41,38,49,47,17,11,1,41,7,16,23,13 }
};
var target = new int[] { 1, 12, 3 };
for (int i = 0; i < target.Length; i++)
{
var sw = Stopwatch.StartNew();
var dummy = new Dummy();
Console.WriteLine("Brute Force answer => {0}, time => {1}", dummy.FindTargetSumWays(ip[i], target[i]), sw.ElapsedMilliseconds);
sw.Restart();
var dp = new DP();
Console.WriteLine("DP with memo answer => {0}, time => {1}", dp.FindTargetSumWays(ip[i], target[i]), sw.ElapsedMilliseconds);
}
#endregion
Console.ReadLine();
}
对此的ouptut是
Nums Called = 1023
Brute Force answer => 256, time => 1
Nums Called = 19
DP with memo answer => 256, time => 1
Nums Called = 2097151
Brute Force answer => 6692, time => 29
Nums Called = 2052849
DP with memo answer => 6692, time => 187
Nums Called = 2097151
Brute Force answer => 5756, time => 28
Nums Called = 2036819
DP with memo answer => 5756, time => 176
我不确定为什么动态方法的时间更加均匀_1186547_方法调用此方法的次数更少 .
2 回答
难怪你的暴力被接受了,因为在最坏的情况下它会是O(2 ^ SizeOfArray) .
在我们的情况下,顺序为2 ^ 20,即约 . 1e6操作的顺序,20是输入中数组大小的上限,如问题中所述 . 如果这个很高,它可能会像DP解决方案那样超时,我们将会看到 .
Coming to the DP solution our recursive relation would be like:
为简单起见,只关注这一部分:
它只取决于以前的状态 . 所以你可以在0-1背包问题中在空间中进行优化,因为问题完全取决于之前的状态 .
运行时复杂度将是
O(2*SizeOfArray*MaxPossibleSum)
,在我们的例子中是O(2 * 20 * 1000),这肯定小于暴力解决方案 . 优化代码的空间复杂度将为O(MaxSum)
.Now regarding problem with your code:
在动态编程中,解决一个大问题应解决许多小问题,这些问题只能解决一次并重复使用多次 . 它被称为overlapping sub-problems属性 . 在这种情况下,您的代码似乎没有利用它 . 为什么?因为在我们的问题中,DP状态由您声明的两个变量“
index
" and "current
”组成,但您只是根据索引输入备忘录 . 这是问题所在 . 我在你的代码中做了一些修改 . 现在它的速度比蛮力的速度快 .我必须说,虽然今天我学到了一点C# . 我之前没有任何经验 .
我不太了解你的memoization代码,但对我来说似乎太复杂了 . 您所需要的只是记住一些数量的nums的所有可能总和的可能组合的数量 . 您一次添加一个数字并更新这些总和 . 你从零和的一个可能的组合开始 .