Updated 28-find-the-index-of-the-first-occurrence-in-a-string.md

gazeldx · gazeldx · commit e7f20e7a370a · 2025-03-22T17:01:16.000+08:00
diff --git a/en/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md b/en/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md
@@ -27,19 +27,31 @@ The first occurrence is at index 0, so we return 0.
 - `haystack` and `needle` consist of only lowercase English characters.
 
 ## Intuition
-Traverse the string once, and if the `needle.length` characters after the current position are equal to `needle`, return the current `index`.
+
+- This kind of question can be solved with one line of code using the built-in `index()`. Obviously, the questioner wants to test our ability to control the loop.
+
+- For `heystack`, traverse each character in turn. There may be two situations:
+    1. First, the character is not equal to the first letter of `needle`. Then process the next character.
+    2. Second, if the character is equal to the first letter of `needle`, continue to compare the next character of `heystack` and `needle` in an internal loop until they are not equal or `needle` has completely matched.
+
+- This question is easier to understand by looking at the code directly.
 
 ## Complexity
-* Time: `O(m * n)`.
+* Time: `O(m + n)`.
 * Space: `O(n)`.
 
 ## Python
 ```python
 class Solution:
     def strStr(self, haystack: str, needle: str) -> int:
         for i in range(len(haystack)):
-            if haystack[i:i + len(needle)] == needle:
-                return i
+            j = 0
+            
+            while i + j < len(haystack) and haystack[i + j] == needle[j]:
+                j += 1
+
+                if j == len(needle):
+                    return i
 
         return -1
 ```
@@ -48,8 +60,14 @@ class Solution:
 ```javascript
 var strStr = function (haystack, needle) {
   for (let i = 0; i < haystack.length; i++) {
-    if (haystack.slice(i, i + needle.length) == needle) {
-      return i
+    let j = 0
+
+    while (i + j < haystack.length && haystack[i + j] == needle[j]) {
+      j += 1
+
+      if (j == needle.length) {
+        return i
+      }
     }
   }
 
diff --git a/zh/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md b/zh/1-1000/28-find-the-index-of-the-first-occurrence-in-a-string.md
@@ -27,7 +27,13 @@
 - `haystack` 和 `needle` 仅由小写英文字符组成
 
 ## 思路
-遍历一遍字符串，如果从当前位置起的后的`needle.length`个字符等于`needle`，则返回当前的`index`。
+
+- 这样的题目，如果用内置的`index()`，一行代码就可以实现。显然，出题人是想考察我们对循环的控制能力。
+- 针对 `heystack`，依次遍历每个字符。可能出现两种情况：
+	1. 该字符与`needle`的首字母不相等。这时处理下一个字符。
+	2. 该字符与`needle`的首字母相等，则在循环中继续比较`heystack`和`needle`的下一个字符，直到不相等或者`needle`已经完全匹配。
+
+- 本题直接看答案比较容易理解。
 
 ## 复杂度
 * 时间：`O(m * n)`。
