[LeetCode]--387. First Unique Character in a String

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.

用数组，简单粗暴，把字符个数都记录下来，最后遍历一下原来的字符串，判断再这个数组里面的个数就行。

public int firstUniqChar1(String s) { if (s == null || s.length() == 0) return -1; int[] count = new int[256]; for (int i = 0; i < s.length(); i++) { int index = s.charAt(i) - 'a'; count[index]++; } for (int i = 0; i < s.length(); i++) { int index = s.charAt(i) - 'a'; if (count[index] == 1) return i; } return -1; }

利用Map来记录。

public int firstUniqChar(String s) { if (s == null || s.length() == 0) { return -1; } HashMap<Character, Integer> hashmap = new HashMap<Character, Integer>(); for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (hashmap.containsKey(c)) { hashmap.put(c, hashmap.get(c) + 1); } else { hashmap.put(c, 1); } } for (int i = 0; i < s.length(); i++) { char c = s.charAt(i); if (hashmap.get(c) == 1) { return i; } } return -1; }

利用一个List一个Map来记录。用更多的空间来换取时间。

public int firstUniqChar(String s) { char[] cs=s.toCharArray(); List<Character> list=new ArrayList<Character>(); Map<Character,Integer> map=new HashMap<Character,Integer>(); for (int i = 0; i < cs.length; i++) { Character c=cs[i]; if (map.containsKey(c)) { list.remove(c); }else{ list.add(c); map.put(c,i); } } return list.size()==0?-1:map.get(list.get(0)); } 相关资源：python入门教程(PDF版)