剑指Offer 50 第一个只出现一次的字符

在一个字符串(0<=字符串长度<=10000，全部由字母组成)中找到第一个只出现一次的字符,并返回它的位置, 如果没有则返回 -1（需要区分大小写）.

 1 # -*- coding:utf-8 -*-
 2 from collections import OrderedDict
 3 class Solution:
 4     def FirstNotRepeatingChar(self, s):
 5         n = len(s)
 6         if n == 0:
 7             return -1
 8         if n == 1:
 9             return -1
10         dic = OrderedDict()
11         for c in s:
12             if c in dic:
13                 dic[c] += 1
14             else:
15                 dic[c] = 1
16         for i in range(n):
17             if dic[s[i]] == 1:
18                 return i
19         return -1
20         # write code here

Java版代码，leetcode地址：

 1 class Solution {
 2     public char firstUniqChar(String s) {
 3         HashMap<Character, Integer> mapTimes = new HashMap<Character, Integer>();
 4         HashMap<Character, Integer> mapIndex = new HashMap<Character, Integer>();
 5         int n = s.length();
 6         char[] ary = s.toCharArray();
 7         for (int i = 0; i < n; i++) {
 8             char c = s.charAt(i);
 9 
10             // 判断是否存在于mapTimes
11             if (mapTimes.containsKey(c)) {
12                 int times = mapTimes.get(c).intValue();
13                 times++;
14                 mapTimes.put(c, times);
15             } else {
16                 mapTimes.put(c, 1);
17             }
18 
19             // 判断是否存在于mapIndex
20             if (!mapIndex.containsKey(c)) {
21                 mapIndex.put(c, i);
22             }
23         }
24         int minIndex = n;
25         for (Character key : mapTimes.keySet()) {
26             if (mapTimes.get(key).intValue() == 1) {
27                 int index = mapIndex.get(key).intValue();
28                 minIndex = Integer.min(minIndex, index);
29             }
30         }
31         if (minIndex == n) {
32             return ' ';
33         } else {
34             return s.charAt(minIndex);
35         }
36     }
37 }

Java第二种写法，使用LinkedHashMap实现：

 1 class Solution {
 2     public char firstUniqChar(String s) {
 3         LinkedHashMap<Character, Integer> map = new LinkedHashMap<>();
 4         int n = s.length();
 5         char[] ary = s.toCharArray();
 6         for (int i = 0; i < n; i++) {
 7             char c = s.charAt(i);
 8 
 9             // 判断是否存在于mapTimes
10             if (map.containsKey(c)) {
11                 int times = map.get(c).intValue();
12                 times++;
13                 map.put(c, times);
14             } else {
15                 map.put(c, 1);
16             }
17         }
18         for(Map.Entry<Character, Integer> entry : map.entrySet()) {
19             if(entry.getValue()==1) {
20                 return entry.getKey();
21             }
22         }
23         return ' ';
24     }
25 }

posted on 2019-06-14 14:07 Sempron2800+ 阅读(132) 评论(0) 编辑收藏举报