387.First Unique Character in a String
Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.
Example 1:
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| Input: s = "leetcode"
Output: 0
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Example 2:
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| Input: s = "loveleetcode"
Output: 2
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Example 3:
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| Input: s = "aabb"
Output: -1
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哈希表存储频数：
时间复杂度：O(n)
空间复杂度：O(S)，S为字符集大小，本题中S不大于26。
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| class Solution {
public:
int firstUniqChar(string s) {
unordered_map<int, int> frequency;
for (char ch: s) {
++frequency[ch];
}
for (int i = 0; i < s.size(); ++i) {
if (frequency[s[i]] == 1) {
return i;
}
}
return -1;
}
};
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哈希表存储索引：
时间复杂度：O(n)
空间复杂度：O(S)，S为字符集大小，本题中S不大于26。
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| class Solution {
public:
int firstUniqChar(string s) {
unordered_map<int, int> position;
int n = s.size();
for (int i = 0; i < n; ++i) {
if (position.count(s[i])) {
position[s[i]] = -1;
}
else {
position[s[i]] = i;
}
}
int first = n;
for (auto [_, pos]: position) {
if (pos != -1 && pos < first) {
first = pos;
}
}
if (first == n) {
first = -1;
}
return first;
}
};
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队列：
时间复杂度：O(n)
空间复杂度：O(S)，S为字符集大小，本题中S不大于26。
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| class Solution {
public:
int firstUniqChar(string s) {
unordered_map<char, int> position;
queue<pair<char, int>> q;
int n = s.size();
for (int i = 0; i < n; ++i) {
if (!position.count(s[i])) {
position[s[i]] = i;
q.emplace(s[i], i);
}
else {
position[s[i]] = -1;
while (!q.empty() && position[q.front().first] == -1) {
q.pop();
}
}
}
return q.empty() ? -1 : q.front().second;
}
};
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