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[LintCode] First Position Unique Character

2024-10-10 12:14:02   213人阅读

Given a string, find the first non-repeating character in it and return it‘s index. If it doesn‘t exist, return -1.

Example

Given s = "lintcode", return 0.

Given s = "lovelintcode", return 2.

 

Solution 1. Two steps‘ iteration 

1. first time iterate through s and create a hash map that maps each unique character to the set of its appearance indices in s.

2. second time iterate through s and return the index of the first character whose apperance set size is 1. return -1 if no such character is found.

 1 public class Solution {
 2     public int firstUniqChar(String s) {
 3         if(s == null || s.length() == 0){
 4             return -1;
 5         }
 6         HashMap<Character, HashSet<Integer>> map = new HashMap<Character, HashSet<Integer>>();
 7         for(int i = 0; i < s.length(); i++){
 8             if(map.containsKey(s.charAt(i))){
 9                 map.get(s.charAt(i)).add(i);
10             }
11             else{
12                 HashSet<Integer> set = new HashSet<Integer>();
13                 set.add(i);
14                 map.put(s.charAt(i),set);
15             }
16         }
17         for(int i = 0; i < s.length(); i++){
18             if(map.get(s.charAt(i)).size() == 1){
19                 return i;
20             }
21         }
22         return -1;
23     }
24 }

 

Optimization on solution 1

Solution 1 already achieves the BCR since we have to iterate through s at least once. 

But we can still optimize on memory usage. In solution 1,  no matter what kind of input 

we get, we always use O(n) space because the hash map stores all characters‘ indices.

Instead of keeping an appearance index set for each unique character, we can simply keep

an integer that counts the appearance time of each unique character. Doing this saves 

memory usage on average cases. In the worst case that each character is unique, we still will

use O(n) memory.

 1 public class Solution {
 2     public int firstUniqChar(String s) {
 3         if(s == null || s.length() == 0){
 4             return -1;
 5         }
 6         HashMap<Character, Integer> map = new HashMap<Character, Integer>();
 7         for(int i = 0; i < s.length(); i++){
 8             if(map.containsKey(s.charAt(i))){
 9                 map.put(s.charAt(i), map.get(s.charAt(i)) + 1);
10             }
11             else{
12                 map.put(s.charAt(i),1);
13             }
14         }
15         for(int i = 0; i < s.length(); i++){
16             if(map.get(s.charAt(i)) == 1){
17                 return i;
18             }
19         }
20         return -1;
21     }
22 }

 

If we can make an assumption about the input string‘s character set is the ASCII set, then we can reduce 

memory usage to O(1).

 1 public class Solution {
 2     public int firstUniqChar(String s) {
 3         int[] alp = new int[256];
 4         char[] arr =s.toCharArray();
 5         for(char c : arr ){
 6             alp[c]++;
 7         }
 8         for(int i = 0; i < arr.length; i++){
 9             if (alp[arr[i]]==1) return i;
10         }
11         return -1;
12     }
13 }

 

 

[LintCode] First Position Unique Character


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