Formatted question description: https://leetcode.ca/all/1065.html
1065. Index Pairs of a String
Level
Easy
Description
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:
- All strings contains only lowercase English letters.
- It’s guaranteed that all strings in
words
are different. 1 <= text.length <= 100
1 <= words.length <= 20
1 <= words[i].length <= 50
- 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).
Solution
For each word
in words
, find all occurrences of word
in text
, and add the index pairs to the result list. Sort the result list and return.
-
class Solution { public int[][] indexPairs(String text, String[] words) { List<int[]> indexPairsList = new ArrayList<int[]>(); for (String word : words) { int wordLength = word.length(); int curIndex = 0; while (curIndex >= 0) { curIndex = text.indexOf(word, curIndex); if (curIndex >= 0) { indexPairsList.add(new int[]{curIndex, curIndex + wordLength - 1}); curIndex++; } } } Collections.sort(indexPairsList, new Comparator<int[]>() { public int compare(int[] array1, int[] array2) { if (array1[0] != array2[0]) return array1[0] - array2[0]; else return array1[1] - array2[1]; } }); int length = indexPairsList.size(); int[][] indexPairs = new int[length][2]; for (int i = 0; i < length; i++) { int[] indexPair = indexPairsList.get(i); indexPairs[i][0] = indexPair[0]; indexPairs[i][1] = indexPair[1]; } return indexPairs; } }
-
Todo
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print("Todo!")