到目标字符串的最短距离

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

/*
 * 思路：
 * 使用Java Stream来简化代码。
 * 我们可以使用IntStream来遍历索引，并计算顺时针和逆时针的距离。
 * 使用filter找到目标字符串的位置，然后使用mapToInt计算距离。
 * 最后使用min找到最小距离。
 */

import java.util.stream.IntStream;

public class Solution {
    public int closetTarget(String[] words, String target, int startIndex) {
        int n = words.length;
        return IntStream.range(0, n)
                .filter(i -> words[i].equals(target))
                .map(i -> {
                    int distanceClockwise = (i - startIndex + n) % n;
                    int distanceCounterClockwise = (startIndex - i + n) % n;
                    return Math.min(distanceClockwise, distanceCounterClockwise);
                })
                .min()
                .orElse(-1);
    }
}

题解首先初始化两个计数器x和y分别表示向右和向左查找目标字符串target的距离。从startIndex开始，首先向右循环遍历words数组，每次移动后增加x的计数。如果索引s超过数组长度，则将其重置为0，模拟环形数组的行为。如果向右的遍历次数x超过数组长度而还未找到目标，则返回-1，表示target不在数组中。完成向右的遍历后，同样的方法向左遍历，每次递减索引s，并增加计数器y。向左遍历时如果索引s变为负数，则重置为数组的最后一个元素，再次模拟环形数组。最后，函数返回x和y中的较小值，即为到达target的最短距离。

O(n)

O(1)