387. 字符串中的第一个唯一字符
public int firstUniqChar(String s) {
Map<Character, Integer> frequency = new HashMap<Character, Integer>();
for (int i = 0; i < s.length(); ++i) {
char ch = s.charAt(i);
frequency.put(ch, frequency.getOrDefault(ch, 0) + 1);
}
for (int i = 0; i < s.length(); ++i) {
if (frequency.get(s.charAt(i)) == 1) {
return i;
}
}
return -1;
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/zi-fu-chuan-zhong-de-di-yi-ge-wei-yi-zi-x9rok/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。public int firstUniqChar(String s) {
Map<Character, Integer> position = new HashMap<Character, Integer>();
Queue<Pair> queue = new LinkedList<Pair>();
int n = s.length();
for (int i = 0; i < n; ++i) {
char ch = s.charAt(i);
if (!position.containsKey(ch)) {
position.put(ch, i);
queue.offer(new Pair(ch, i));
} else {
position.put(ch, -1);//将队列头部重复的数据移除
while (!queue.isEmpty() && position.get(queue.peek().ch) == -1) {
queue.poll();
}
}
}
return queue.isEmpty() ? -1 : queue.poll().pos;
}
class Pair {
char ch;
int pos;
Pair(char ch, int pos) {
this.ch = ch;
this.pos = pos;
}
}
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/zi-fu-chuan-zhong-de-di-yi-ge-wei-yi-zi-x9rok/解析
小贴士
在维护队列时，我们使用了「延迟删除」这一技巧。也就是说，即使队列中有一些字符出现了超过一次，但它只要不位于队首，那么就不会对答案造成影响，我们也就可以不用去删除它。只有当它前面的所有字符被移出队列，它成为队首时，我们才需要将它移除。
链接：https://leetcode-cn.com/problems/first-unique-character-in-a-string/solution/zi-fu-chuan-zhong-de-di-yi-ge-wei-yi-zi-x9rok/
