字符串中的第一个唯一字符

From process-oriented thinking to object-oriented thinking

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.

Example

s = "leetcode"
return 0.

s = "loveleetcode",
return 2.

Note:

You may assume the string contain only lowercase letters.

我的（ 2 ms ）

public int firstUniqChar(String s) {
		int len = s.length();
		for (char k = 'a'; k <= 'z'; k++) {
			int p1 = s.indexOf(k);// -1 不存在
			int p2 = s.lastIndexOf(k);
			if (p1 == p2 && p1 != -1) {
				len = Math.min(len, p1);
			}
		}
		if (len == s.length()) {
			return -1;
		}
		return len;
    }
}

从字母而不是数组开始扫描。

官方

public int firstUniqChar(String s) {
		HashMap<Character, Integer> map = new HashMap<Character, Integer>();
		int n = s.length();
				
	         for(int i = 0; i<n; i++){
		             char c = s.charAt(i);
		             map.put(c, map.getOrDefault(c,0)+1);
		         }
				
		         for(int j = 0; j<n;j++){
		             if(map.get(s.charAt(j)) == 1){
		                 return j;
		             }
		         }	
		         return -1;
    }