[Swift]LeetCode245.最短单词距离 III $ Shortest Word Distance III

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This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

For example,
Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Given word1 = “makes”, word2 = “coding”, return 1.
Given word1 = "makes", word2 = "makes", return 3.

Note:
You may assume word1 and word2 are both in the list.

---

这是最短单词距离的后续行动。现在唯一的区别是word1可以与word2相同。

给定单词列表以及单词1和单词2，返回列表中这两个单词之间的最短距离。

word1和word2可能相同，它们代表列表中的两个单独单词。

例如，

假设words=[“practice”、“makes”、“perfect”、“coding”、“makes”]。

给定word1=“makes”，word2=“coding”，返回1。

给定word1=“makes”，word2=“makes”，返回3。

注：

您可以假定word1和word2都在列表中。

---

class Solution {
    func shortestWordDistance(_ words: [String],_ word1:String,_ word2:String) -> Int {
        var idx:Int = -1
        var res:Int = Int.max
        for i in 0..<words.count
        {
            if words[i] == word1 || words[i] == word2
            {
                if idx != -1 && (word1 == word2 || words[i] != words[idx])
                {
                    res = min(res, i - idx)
                }
                idx = i
            }
        }
        return res
    }
}