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387. First Unique Character in a String
还是要善用 Map 来统计字符数量！
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.
思路分析
通过解法是再次遍历字符串，思考如何遍历字母表来获取第一个唯一字符的位置。
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/**
* Runtime: 34 ms, faster than 54.19% of Java online submissions for First Unique Character in a String.
*
* Memory Usage: 37.4 MB, less than 100.00% of Java online submissions for First Unique Character in a String.
*
* Copy from: https://leetcode.com/problems/first-unique-character-in-a-string/solution/[First Unique Character in a String solution - LeetCode]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-11 10:48
*/
public int firstUniqChar(String s) {
if (Objects.isNull(s) || s.length() == 0) {
return -1;
}
char[] chars = s.toCharArray();
Map<Character, Integer> charToCountMap = new HashMap<>();
for (char aChar : chars) {
Integer count = charToCountMap.getOrDefault(aChar, 0);
charToCountMap.put(aChar, ++count);
}
for (int i = 0; i < chars.length; i++) {
if (charToCountMap.get(chars[i]) == 1) {
return i;
}
}
return -1;
}
/**
* Runtime: 53 ms, faster than 10.48% of Java online submissions for First Unique Character in a String.
*
* Memory Usage: 38.4 MB, less than 96.43% of Java online submissions for First Unique Character in a String.
*/
public int firstUniqCharArray(String s) {
if (Objects.isNull(s) || s.length() == 0) {
return -1;
}
char[] chars = s.toCharArray();
boolean[] unique = new boolean[chars.length];
Arrays.fill(unique, true);
for (int i = 0; i < chars.length; i++) {
if (!unique[i]) {
continue;
}
for (int j = i + 1; j < chars.length; j++) {
if (!unique[j]) {
continue;
}
if (chars[i] == chars[j]) {
unique[i] = false;
unique[j] = false;
}
}
}
for (int i = 0; i < unique.length; i++) {
if (unique[i]) {
return i;
}
}
return -1;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-11 10:48
*/
public int firstUniqChar(String s) {
int[] cnt = new int[26];
for (int i = 0; i < s.length(); i++) {
int idx = s.charAt(i) - 'a';
cnt[idx]++;
}
for (int i = 0; i < s.length(); i++) {
int idx = s.charAt(i) - 'a';
if (cnt[idx] == 1) {
return i;
}
}
return -1;
}