opt: opt 028

Author: Blankj

note/028/README.md

@@ -11,18 +11,20 @@ Returns the index of the first occurrence of needle in haystack, or -1 if needle
 
 ## 思路
 
-题意是从主串中找到子串的索引，如果找不到则返回-1，我们只需要遍历主串长度减子串长度即可，利用substring比较即可。
+题意是从主串中找到子串的索引，如果找不到则返回-1，当字串长度大于主串，直接返回-1，然后我们只需要遍历比较即可。
 
 ```java
 class Solution {
     public int strStr(String haystack, String needle) {
-        int l1 = haystack.length(), l2 = needle.length(), l3 = l1 - l2;
-        for (int i = 0; i <= l3; ++i) {
-            if (haystack.substring(i, i + l2).equals(needle)) {
-                return i;
+        int l1 = haystack.length(), l2 = needle.length();
+        if (l1 < l2) return -1;
+        for (int i = 0; ; i++) {
+            for (int j = 0; ; j++) {
+                if (j == l2) return i;
+                if (i + j == l1) return -1;
+                if (haystack.charAt(i + j) != needle.charAt(j)) break;
             }
         }
-        return -1;
     }
 }
 ```

src/com/blankj/easy/_028/Solution.java

@@ -10,13 +10,15 @@
  */
 public class Solution {
     public int strStr(String haystack, String needle) {
-        int l1 = haystack.length(), l2 = needle.length(), l3 = l1 - l2;
-        for (int i = 0; i <= l3; ++i) {
-            if (haystack.substring(i, i + l2).equals(needle)) {
-                return i;
+        int l1 = haystack.length(), l2 = needle.length();
+        if (l1 < l2) return -1;
+        for (int i = 0; ; i++) {
+            for (int j = 0; ; j++) {
+                if (j == l2) return i;
+                if (i + j == l1) return -1;
+                if (haystack.charAt(i + j) != needle.charAt(j)) break;
             }
         }
-        return -1;
     }
 
     public static void main(String[] args) {