From 2c41ea320fc8d65e8b2501a611d9ca2fbe75f6e1 Mon Sep 17 00:00:00 2001
From: kingeasternsun <kingeasternsun@gmail.com>
Date: Wed, 23 Dec 2020 12:53:01 +0800
Subject: [PATCH] Update 0387.First-Unique-Character-in-a-String.md

---
 ...0387.First-Unique-Character-in-a-String.md | 55 ++++++++++++++++++-
 1 file changed, 54 insertions(+), 1 deletion(-)

diff --git a/website/content/ChapterFour/0387.First-Unique-Character-in-a-String.md b/website/content/ChapterFour/0387.First-Unique-Character-in-a-String.md
index b3ccff4c..6f65a982 100755
--- a/website/content/ChapterFour/0387.First-Unique-Character-in-a-String.md
+++ b/website/content/ChapterFour/0387.First-Unique-Character-in-a-String.md
@@ -45,4 +45,57 @@ func firstUniqChar(s string) int {
 	return -1
 }
 
-```
\ No newline at end of file
+```
+
+## 思路2
+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
+
+## 代码
+
+执行用时: 8 ms
+内存消耗: 5.2 MB
+
+
+```go
+
+
+func firstUniqChar(s string) int {
+
+	charMap := make([][2]int, 26)
+	for i := 0; i < 26; i++ {
+		charMap[i][0] = -1
+		charMap[i][1] = -1
+	}
+
+	for i := 0; i < len(s); i++ {
+		if charMap[s[i]-'a'][0] == -1 {
+			charMap[s[i]-'a'][0] = i
+		} else { //已经出现过
+			charMap[s[i]-'a'][1] = i
+		}
+
+	}
+
+	res := len(s)
+
+	for i := 0; i < 26; i++ {
+
+		//只出现了一次
+		if charMap[i][0] >= 0 && charMap[i][1] == -1 {
+			if charMap[i][0] < res {
+				res = charMap[i][0]
+			}
+		}
+
+	}
+
+	if res == len(s) {
+		return -1
+	}
+	return res
+}
+
+```