Swift 无法解开可选内容

Question

这个问题已经有答案了:

奥 git _a (17 个回答)

已关闭 7 年前。

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples: pattern = "abba", str = "dog cat cat dog" should return true.

pattern = "abba", str = "dog cat cat fish" should return false.

pattern = "aaaa", str = "dog cat cat dog" should return false.

pattern = "abba", str = "dog dog dog dog" should return false.

Notes: You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

class Solution
{
    func wordPattern(pattern: String, _ str: String) -> Bool
    {

        let arr = str.characters.split{$0 == " "}.map(String.init)
        if arr.count != pattern.characters.count
        {
            return false
        }

        var dict = [Character:String]()

        for i in (0...arr.count-1)
        {
            let cha = pattern[pattern.startIndex.advancedBy(i)]

            if dict.keys.contains(cha)
            {
                if dict[cha]! != arr[i]
                {
                    return false
                }
            }
            else
            {
                if dict.values.contains(dict[cha]!)
                //fatal error: unexpectedly found nil while unwrapping an Optional value
                {
                    return false
                }

                dict[cha] = arr[i]
            }

        }
        return true
    }
}


var test = Solution()
var result = test.wordPattern("abba", "dog cat cat dog")
print(result)

I do not know "fatal error: unexpectedly found nil while unwrapping an Optional value". Any help, I appreciate it. Thank you so much!

Answer 1

修改您的条件。

当 dict.keys.contains(cha) 为 false 时，换句话说，当字典不存在时，将执行您的 else 分支包含cha。然后，您尝试访问您已经知道必须返回 nil 的 dict[cha] (因为 cha 不在字典中)。然后你尝试打开它。展开 nil 会使您的应用程序崩溃。

还要考虑该条件正在尝试检查从字典中获取的值是否包含在字典中。你想做什么？

您想要的可能只是(注意，没有令人讨厌的 !):

let arrayValue = arr[i]

if let value = dict[cha] {
    if value != arrayValue {
        return false
    }
} else {
    if dict.values.contains(arrayValue) { // maybe this? it's hard to guess
        return false
    }

    dict[cha] = arrayValue
}