标签：etc ati character cte 时间 interview content lower map
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.
一开始用hashmap做的，很慢。
public class Solution {
public int firstUniqChar(String s) {
if (s == null) {
return -1;
}
Map<Character, Integer> hm = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
if (hm.containsKey(s.charAt(i))) {
hm.put(s.charAt(i), hm.get(s.charAt(i)) + 1);
} else {
hm.put(s.charAt(i), 0);
}
}
for (int i = 0; i < s.length(); i++) {
if (hm.get(s.charAt(i)) == 0) {
return i;
}
}
return -1;
}
}
后来发现可以用数组做，就很快了。记得应该做过，怎么就不记得了呢。。。
public class Solution {
public int firstUniqChar(String s) {
if (s == null) {
return -1;
}
int[] f = new int[26];
for (int i = 0; i < s.length(); i++) {
f[s.charAt(i) - ‘a‘]++;
}
for (int i = 0; i < s.length(); i++) {
if (f[s.charAt(i) - ‘a‘] == 1) {
return i;
}
}
return -1;
}
}
其实不是很懂为什么时间复杂度都一样，却从120多ms变成30ms。
附上C++做法：
class Solution {
public:
int firstUniqChar(string s) {
unordered_map<char, int> m;
for (auto &c : s) {
m[c]++;
}
for (int i = 0; i < s.size(); i++) {
if (m[s[i]] == 1) {
return i;
}
}
return -1;
}
};
如果string过长的话，就遍历hashmap而不是string
class Solution {
public:
int firstUniqChar(string s) {
unordered_map<char, pair<int, int>> m;
int index = s.size();
for (int i = 0; i < s.size(); i++) {
m[s[i]].first++;
m[s[i]].second = i;
}
for (auto &p : m) {
if (p.second.first == 1) {
index = min(index, p.second.second);
}
}
return index == s.size() ? -1 : index;
}
};
class Solution {
public:
int firstUniqChar(string s) {
int f[26] = {0};
for (int i = 0; i < s.size(); i++) {
f[s[i] - ‘a‘]++;
}
for (int i = 0; i < s.size(); i++) {
if (f[s[i] - ‘a‘] == 1) {
return i;
}
}
return -1;
}
};
First Unique Character in a String
标签：etc ati character cte 时间 interview content lower map
原文地址：http://www.cnblogs.com/aprilyang/p/6984082.html
