QUESTION:
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] is words[(i + 1) % n] and the previous element of words[i] is words[(i - 1 + n) % n], where n is the length of words. 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] and target consist of only lowercase English letters.
0 <= startIndex < words.length
EXPLANATION:
其实就主要分成3种情况, 一种是在同一个方向上. 另外一个就比较复杂. 得分成两种情况, 一种是startindex在前, target在后, 一种是target在前, startindex在后. 所以只要将这三种情况都对比上就可以了.
SOLUTION:
class Solution {
func closetTarget(_ words: [String], _ target: String, _ startIndex: Int) -> Int {
if !words.contains(target) {
return -1
}
var result = Int.max
for (i, word) in words.enumerated() {
if word == target {
result = min(result, abs(startIndex - i)) // 在同一个圈里
result = min(result, words.count - startIndex + i)
result = min(result, words.count - i + startIndex)
}
}
return result
}
}