1、题目
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.
2、代码实现
public class Solution {
public int firstUniqChar(String s) {
if (s == null || s.length() == 0) {
return -1;
}
HashMap<Character, Integer> map = new HashMap<Character, Integer>();
for (int i = 0; i < s.length(); i++) {
Integer in = map.get(s.charAt(i));
if (in == null)
map.put(s.charAt(i), 1);
else
map.put(s.charAt(i), 2);
}
for (int i = 0; i < s.length(); i++) {
if(map.get(s.charAt(i)) == 2) {
continue;
} else {
if (map.get(s.charAt(i)) == 1) {
return i;
}
}
}
return -1;
}
}
3、总结
一般看到求数组里面唯一元素，和字符串里面唯一元素，我们可以通过HashMap来解决，每个字符或者元素作为key,然后出现一次设置一个value1,出现2次以上设置一个统一的value2,最后通过遍历得到value1,来解决问题
