Arrays 如何在Swift中识别数组中特定值的连续出现？

我需要评估一个数组是否有两个连续出现的特定值。例如，我想评估一个数组在连续索引位置是否有两个值“1”的实例。我只希望对于值“1”的重复（即，不适用于其他重复值）和仅适用于连续索引位置的重复，该值计算为“真”。我发现了许多模糊不清的类似问题和解决方案，但没有一个与这种特定情况相匹配。具体例子如下：

    [0,1,1,0,2,0]  // Evaluates to True Because of Two Consecutive 1's
    [0,1,0,1,2,2]  // Evaluates to False, In Spite of Repeated 2's and Multiple (Non-Consecutive) Instances of 1

使用：

let specificValue: Int = 1
let numberArray: [Int] = [0,1,1,0,2,0]
let hasConsecutiveValue = checkForConsecutiveValue(value: specificValue, within: numberArray)

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您可以使用for循环来实现这一点。我在

序列

协议中做了一些扩展，使解决方案更加通用：

extension Sequence where Element: Equatable {
    func consecutiveOccurrences(of element: Element) -> Bool {
        var appeared = false
        for item in self {
            if item == element, appeared {
                return true
            } else if item == element {
                appeared = true
            } else {
                appeared = false
            }
        }
        return false
    }
}

print([0,1,1,0,2,0].consecutiveOccurrences(of: 1)) // true
print([0,1,0,1,2,2].consecutiveOccurrences(of: 1)) // false

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@gcharita kkkk我们让他很难过。他至少展示了一些effort@Eric33187

用于numberraray.index.dropLast（）中的索引{

谢谢你，Leo.lol.我添加了一个保护声明。现在大家都高兴了吗？哈哈哈，如果用户拥有这个特权（足够的声誉）是的。但是如果你编辑自己的PostOk更有意义，那就更好了。谢谢你的提示。也谢谢所有的指针大家！一个简单的

for

循环，比较当前值和上一个（或下一个）值怎么样？为什么这个如此特别，以至于你还没有尝试过那个简单的解决方案？

extension Sequence where Element: Equatable {
    func consecutiveOccurrences(of element: Element) -> Bool {
        var appeared = false
        for item in self {
            if item == element, appeared {
                return true
            } else if item == element {
                appeared = true
            } else {
                appeared = false
            }
        }
        return false
    }
}

print([0,1,1,0,2,0].consecutiveOccurrences(of: 1)) // true
print([0,1,0,1,2,2].consecutiveOccurrences(of: 1)) // false