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81 lines
1.8 KiB
Markdown
Executable File
81 lines
1.8 KiB
Markdown
Executable File
# [387. First Unique Character in a String](https://leetcode.com/problems/first-unique-character-in-a-string/)
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## 题目
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Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1.
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**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.
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**Note**: You may assume the string contain only lowercase letters.
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## 题目大意
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给定一个字符串，找到它的第一个不重复的字符，并返回它的索引。如果不存在，则返回 -1。
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## 解题思路
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- 简单题，要求输出第一个没有重复的字符。
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- 解法二这个思路只超过 81% 的用户，但是如果测试样例中 s 的字符串很长，但是满足条件的字符都在靠后的位置的话，这个思路应该会更有优势。通过记录每个字符的第一次出现的位置和最后一次出现的位置。第一次对 s 进行一次遍历。第二次仅仅对数组进行遍历就可以了。
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## 代码
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```go
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package leetcode
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// 解法 一
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func firstUniqChar(s string) int {
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result := make([]int, 26)
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for i := 0; i < len(s); i++ {
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result[s[i]-'a']++
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}
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for i := 0; i < len(s); i++ {
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if result[s[i]-'a'] == 1 {
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return i
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}
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}
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return -1
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}
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// 解法 二
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// 执行用时: 8 ms
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// 内存消耗: 5.2 MB
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func firstUniqChar1(s string) int {
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charMap := make([][2]int, 26)
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for i := 0; i < 26; i++ {
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charMap[i][0] = -1
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charMap[i][1] = -1
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}
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for i := 0; i < len(s); i++ {
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if charMap[s[i]-'a'][0] == -1 {
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charMap[s[i]-'a'][0] = i
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} else { //已经出现过
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charMap[s[i]-'a'][1] = i
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}
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}
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res := len(s)
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for i := 0; i < 26; i++ {
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//只出现了一次
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if charMap[i][0] >= 0 && charMap[i][1] == -1 {
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if charMap[i][0] < res {
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res = charMap[i][0]
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}
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}
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}
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if res == len(s) {
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return -1
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}
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return res
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}
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``` |