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;
}
};
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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);
}
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;
}
}
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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;
}
}
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