[Swift]LeetCode1065. 字符串的索引对 | Index Pairs of a String

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Given a text string and words (a list of strings), return all index pairs [i, j] so that the substring text[i]...text[j] is in the list of words.

Example 1:

Input: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
Output: [[3,7],[9,13],[10,17]]

Example 2:

Input: text = "ababa", words = ["aba","ab"]
Output: [[0,1],[0,2],[2,3],[2,4]]
Explanation: 
Notice that matches can overlap, see "aba" is found in [0,2] and [2,4]. 

Note:

1. All strings contains only lowercase English letters.
2. It's guaranteed that all strings in words are different.
3. 1 <= text.length <= 100
4. 1 <= words.length <= 20
5. 1 <= words[i].length <= 50
6. Return the pairs [i,j] in sorted order (i.e. sort them by their first coordinate in case of ties sort them by their second coordinate).

---

给出 字符串 text 和 字符串列表 words, 返回所有的索引对 [i, j] 使得在索引对范围内的子字符串 text[i]...text[j]（包括 i 和 j）属于字符串列表 words。
示例 1:

输入: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
输出: [[3,7],[9,13],[10,17]]
示例 2:

输入: text = "ababa", words = ["aba","ab"]
输出: [[0,1],[0,2],[2,3],[2,4]]
解释:
注意，返回的配对可以有交叉，比如，"aba" 既在 [0,2] 中也在 [2,4] 中
提示:

　1.所有字符串都只包含小写字母。
　　2.保证 words 中的字符串无重复。
　　3.1 <= text.length <= 100
　　4.1 <= words.length <= 20
　　5.1 <= words[i].length <= 50
　　6.按序返回索引对 [i,j]（即，按照索引对的第一个索引进行排序，当第一个索引对相同时按照第二个索引对排序）。

---

Runtime: 92 ms

Memory Usage: 21.4 MB

class Solution {
    func indexPairs(_ text: String, _ words: [String]) -> [[Int]] {
        var ret:[[Int]] = [[Int]]()
        let arrText:[Character] = Array(text)
        for word in words
        {
            let arrStr:[Character] = Array(word)
            let num:Int = arrStr.count
            for i in 0..<arrText.count
            {
                if (arrText[i] == arrStr.first!) && (i + num <= arrText.count) && (Array(arrText[i..<(i + num)]) == arrStr)
                {
                    ret.append([i,i + num - 1])
                }
            }
        }
        ret.sort(){
            if $0[0] == $1[0]
            {
                return $0[1] <= $1[1]
            }
            else
            {
                return $0[0] <= $1[0]
            }
        }
        return ret
    }
}