Class Solution
Easy
You are given a 0-indexed circular string array words
and a string target
. A circular array means that the array’s end connects to the array’s beginning.
- Formally, the next element of
words[i]
iswords[(i + 1) % n]
and the previous element ofwords[i]
iswords[(i - 1 + n) % n]
, wheren
is the length ofwords
.
Starting from startIndex
, you can move to either the next word or the previous word with 1
step at a time.
Return the shortest distance needed to reach the string target
. If the string target
does not exist in words
, return -1
.
Example 1:
Input: words = [“hello”,“i”,“am”,“leetcode”,“hello”], target = “hello”, startIndex = 1
Output: 1
Explanation: We start from index 1 and can reach “hello” by
-
moving 3 units to the right to reach index 4.
-
moving 2 units to the left to reach index 4.
-
moving 4 units to the right to reach index 0.
-
moving 1 unit to the left to reach index 0. The shortest distance to reach “hello” is 1.
Example 2:
Input: words = [“a”,“b”,“leetcode”], target = “leetcode”, startIndex = 0
Output: 1
Explanation: We start from index 0 and can reach “leetcode” by
-
moving 2 units to the right to reach index 3.
-
moving 1 unit to the left to reach index 3.
The shortest distance to reach “leetcode” is 1.
Example 3:
Input: words = [“i”,“eat”,“leetcode”], target = “ate”, startIndex = 0
Output: -1
Explanation: Since “ate” does not exist in words
, we return -1.
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 100
words[i]
andtarget
consist of only lowercase English letters.0 <= startIndex < words.length
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Constructor Summary
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Method Summary
Modifier and TypeMethodDescriptionint
closetTarget
(java.lang.String[] words, java.lang.String target, int startIndex) Methods inherited from class java.lang.Object
clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait
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Constructor Details
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Solution
public Solution()
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Method Details
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closetTarget
public int closetTarget(java.lang.String[] words, java.lang.String target, int startIndex)
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