leetcode解题: First Unique Character in a String (387)

Given a string, find the first non-repeating character in it and return it’s index. If it doesn’t exist, return -1.

Examples:

s = “leetcode”
return 0.

s = “loveleetcode”,
return 2.
Note: You may assume the string contain only lowercase letters.

解法1：HashTable

思路是将字符串建立hashtable，其中存的是每一个字符出现的位置，如果出现超过1次则位置设置为-1，然后遍历hashtable找出最小的非-1的即可。
C++

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class { public: int firstUniqChar(string s) { unordered_map<char, int> map; for (int i = 0; i < s.size(); ++i) { if (map.find(s[i]) != map.end()) { map[s[i]] = -1; } else { map[s[i]] = i; } } int res = s.size(); for (auto iter = map.begin(); iter != map.end(); ++iter) { if (iter->second !=-1) { res = min(res, iter->second); } } if (res == s.size()) return -1; else return res; } };

Java

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class { public int firstUniqChar(String s) { if (s == null || s.length() == 0) { return -1; } Map<Character, List<Integer>> map = new HashMap<>(); for (int i = 0; i < s.length(); i++) { char ch = s.charAt(i); if (!map.containsKey(ch)) { map.put(ch, new ArrayList<Integer>()); } map.get(ch).add(i); } 大专栏  leetcode解题: First Unique Character in a String (387)ne"> int res = Integer.MAX_VALUE; for (char key : map.keySet()) { if (map.get(key).size() == 1) { res = Math.min(res, map.get(key).get(0)); } } return res == Integer.MAX_VALUE ? -1 : res; } }

解法2：HashTable

上面的思路有点繁复，如果hashtable中存入每一个字符出现的次数，那么只需要重新扫描一遍字符串，找到第一个次数为1的就是所求的答案。
C++

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class { public: int firstUniqChar(string s) { unordered_map<char, int> map; for (char c : s) { ++map[c]; } for (int i = 0; i < s.size(); i++) { if (map[s[i]] == 1) { return i; } } return -1; } };

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class { public int firstUniqChar(String s) { if (s == null || s.length() == 0) { return -1; } Map<Character, Integer> map = new HashMap<>(); for (int i = 0; i < s.length(); i++) { char ch = s.charAt(i); map.put(ch, map.getOrDefault(ch, 0) + 1); } for (int i = 0; i < s.length(); i++) { char ch = s.charAt(i); if (map.get(ch) == 1) { return i; } } return -1; } }