Class Solution
java.lang.Object
g0801_0900.s0821_shortest_distance_to_a_character.Solution
821 - Shortest Distance to a Character.<p>Easy</p>
<p>Given a string <code>s</code> and a character <code>c</code> that occurs in <code>s</code>, return <em>an array of integers</em> <code>answer</code> <em>where</em> <code>answer.length == s.length</code> <em>and</em> <code>answer[i]</code> <em>is the <strong>distance</strong> from index</em> <code>i</code> <em>to the <strong>closest</strong> occurrence of character</em> <code>c</code> <em>in</em> <code>s</code>.</p>
<p>The <strong>distance</strong> between two indices <code>i</code> and <code>j</code> is <code>abs(i - j)</code>, where <code>abs</code> is the absolute value function.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> s = “loveleetcode”, c = “e”</p>
<p><strong>Output:</strong> [3,2,1,0,1,0,0,1,2,2,1,0]</p>
<p><strong>Explanation:</strong> The character ‘e’ appears at indices 3, 5, 6, and 11 (0-indexed).</p>
<p>The closest occurrence of ‘e’ for index 0 is at index 3, so the distance is abs(0 - 3) = 3.</p>
<p>The closest occurrence of ‘e’ for index 1 is at index 3, so the distance is abs(1 - 3) = 2.</p>
<p>For index 4, there is a tie between the ‘e’ at index 3 and the ‘e’ at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.</p>
<p>The closest occurrence of ‘e’ for index 8 is at index 6, so the distance is abs(8 - 6) = 2.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> s = “aaab”, c = “b”</p>
<p><strong>Output:</strong> [3,2,1,0]</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= s.length <= 10<sup>4</sup></code></li>
<li><code>s[i]</code> and <code>c</code> are lowercase English letters.</li>
<li>It is guaranteed that <code>c</code> occurs at least once in <code>s</code>.</li>
</ul>
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Constructor Summary
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Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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shortestToChar
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