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This article is about how LeetCode finds the first unique character in a string. The editor thinks it is very practical, so share it with you as a reference and follow the editor to have a look.

1. Brief introduction of the problem.

Given a string, find its first non-repeating character and return its index.

If it does not exist, return-1.

2, example

Example:

S = "leetcode" returns 0

S = "loveleetcode" returns 2

Tip: you can assume that the string contains only lowercase letters.

3, the train of thought of solving the problem

The use of keys and values on the collection LinkedHashMap

4, problem solving procedure

Import java.util.HashMap;import java.util.LinkedHashMap;import java.util.Map;import java.util.Optional

Public class FirstUniquCharTest2 {public static void main (String [] args) {String s = "cc"; int firstUniqChar = firstUniqChar (s); System.out.println ("firstUniqChar =" + firstUniqChar);}

Public static int firstUniqChar (String s) {if (s = = null | | s.length () = = 0) {return-1;} char [] toCharArray = s.toCharArray (); HashMap hashMap = new LinkedHashMap (toCharArray.length); for (char c: toCharArray) {hashMap.put (c, hashMap.getOrDefault (c, 0) + 1) } Optional optionalEntry = hashMap.entrySet (). Stream (). Filter (x-> x.getValue () = = 1). FindFirst (); if (! optionalEntry.isPresent ()) {return-1;} else {Character character = optionalEntry.get (). GetKey (); return s.indexOf (character);}

}}

5. Picture version of the problem solving program.

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