LeetCode 387. First Unique Character in a String (Java+C/C++)

题意：输出字符串中唯一出现的第一个字符的索引下标

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.

Java代码：

public class Solution {
public int firstUniqChar(String s) {
char[] arr = s.toCharArray();
int[] cache = new int[26];
for(int i = 0; i < arr.length; i++) {
++cache[arr[i] - 'a'];
}
for(int i = 0; i < arr.length; i++) {
if(cache[arr[i] - 'a'] == 1)
return i;
}
return -1;
}
}

C++代码：

class Solution {
public:
int firstUniqChar(string s) {
int cache[26] = {0};
for(int i = 0; i < s.length(); i++) {
++cache[s[i] - 'a'];
}
for(int i = 0; i < s.length(); i++) {
if(cache[s[i] - 'a'] == 1)
return i;
}
return -1;
}
};

C代码：

int firstUniqChar(char* s) {
int len = strlen(s);
int cache[26] = {0};
for(int i = 0; i < len; i++) {
++cache[s[i] - 'a'];
}
for(int i = 0; i < len; i++) {
if(cache[s[i] - 'a'] == 1)
return i;
}
return -1;
}

解题思路：很显然，需要一个缓存存储每个字符出现的次数；由于题目说明只出现小写字母，所以可以开辟一个26长度的数组存储每个字母出现的次数，遍历两次，具体看代码，时间复杂度O(n)