Sub*_*eep 5 arrays algorithm data-structures
现在可用的解决方案是每个地方都有一个include和exclude sum 。在max这两个结束时会给我输出。
现在最初我很难理解这个算法,我想为什么不以简单的方式进行。
算法:通过一次增加两个数组指针来循环数组
sumsum最后,拿max这两个sum。
那样的话,我认为复杂度会减半 O(n/2)
这个算法正确吗?
这是动态规划的一个例子。算法是:
让我们展示第二步,假设我们有:
[1, 2, 3, 4, 5, 6, 10, 125, -8, 9]
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1是积极的,这就是为什么
take_sum = max(1 + max_sum([3, 4, 5, 6, 10, 125, -8, 9])) // we take "1"
skip_sum = max_sum([2, 3, 4, 5, 6, 10, 125, -8, 9]) // we skip "1"
max_sum = max(take_sum, skip_sum)
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C#实现(最简单的代码,为了展示赤裸裸的想法,不再进一步优化):
private static int BestSum(int[] array, int index) {
if (index >= array.Length)
return 0;
if (array[index] <= 0)
return BestSum(array, index + 1);
int take = array[index] + BestSum(array, index + 2);
int skip = BestSum(array, index + 1);
return Math.Max(take, skip);
}
private static int BestSum(int[] array) {
return BestSum(array, 0);
}
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测试:
Console.WriteLine(BestSum(new int[] { 1, -2, -3, 100 }));
Console.WriteLine(BestSum(new int[] { 100, 8, 10, 20, 7 }))
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结果:
101
120
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请检查您的初始算法是否返回98以及117哪些是次优总和。
编辑:在现实生活中,您可能想要添加一些优化,例如记忆和特殊情况测试:
private static Dictionary<int, int> s_Memo = new Dictionary<int, int>();
private static int BestSum(int[] array, int index) {
if (index >= array.Length)
return 0;
int result;
if (s_Memo.TryGetValue(index, out result)) // <- Memoization
return result;
if (array[index] <= 0)
return BestSum(array, index + 1);
// Always take, when the last item to choose or when followed by non-positive item
if (index >= array.Length - 1 || array[index + 1] <= 0) {
result = array[index] + BestSum(array, index + 2);
}
else {
int take = array[index] + BestSum(array, index + 2);
int skip = BestSum(array, index + 1);
result = Math.Max(take, skip);
}
s_Memo.Add(index, result); // <- Memoization
return result;
}
private static int BestSum(int[] array) {
s_Memo.Clear();
return BestSum(array, 0);
}
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测试:
using System.Linq;
...
Random gen = new Random(0); // 0 - random, by repeatable (to reproduce the same result)
int[] test = Enumerable
.Range(1, 10000)
.Select(i => gen.Next(100))
.ToArray();
int evenSum = test.Where((v, i) => i % 2 == 0).Sum();
int oddSum = test.Where((v, i) => i % 2 != 0).Sum();
int suboptimalSum = Math.Max(evenSum, oddSum); // <- Your initial algorithm
int result = BestSum(test);
Console.WriteLine(
$"odd: {oddSum} even: {evenSum} suboptimal: {suboptimalSum} actual: {result}");
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结果:
odd: 246117 even: 247137 suboptimal: 247137 actual: 290856
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