LeetCode之First Unique Character in a String
【摘要】 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 ...
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:
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s = "leetcode"
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return 0.
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s = "loveleetcode",
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return 2.
2、代码实现
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public class Solution {
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public int firstUniqChar(String s) {
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if (s == null || s.length() == 0) {
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return -1;
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}
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HashMap<Character, Integer> map = new HashMap<Character, Integer>();
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for (int i = 0; i < s.length(); i++) {
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Integer in = map.get(s.charAt(i));
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if (in == null)
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map.put(s.charAt(i), 1);
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else
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map.put(s.charAt(i), 2);
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}
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for (int i = 0; i < s.length(); i++) {
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if(map.get(s.charAt(i)) == 2) {
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continue;
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} else {
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if (map.get(s.charAt(i)) == 1) {
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return i;
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}
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}
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}
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return -1;
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}
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}
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
文章来源: chenyu.blog.csdn.net,作者:chen.yu,版权归原作者所有,如需转载,请联系作者。
原文链接:chenyu.blog.csdn.net/article/details/67725141
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