fix: 优化removeItem和isPathMatched

Author: zhangfisher

packages/native/src/utils/isPathMatched.ts

@@ -1,40 +1,67 @@
-
 /**
- * 
+ *
  * 判断path是否与pattern匹配
- * 
+ *
  * isPathMatched("a.b.c","a.b.c")  == true
  * isPathMatched("a.b.c","a.b.*")  == true
  * isPathMatched("a.b.c","a.*.*")  == true
  * isPathMatched("a.b.c","*.*.*")  == true
  * isPathMatched("a.b.c",".b.*")  == true
  * isPathMatched("a.b.c.d","a.**")  == true
- * 
+ *
  * - '**' 匹配后续的
  * - '*' 匹配任意数量的字符，包括零个字符
- * 
- * @param path 
- * @param pattern 
+ *
+ * @param path
+ * @param pattern
  */
-export function isPathMatched(path:string[],pattern:string[]):boolean{
-    if(path.length !== pattern.length && (path.length>0 && pattern[pattern.length-1]!=='**') ){
+export function isPathMatched(path: string[], pattern: string[]): boolean {
+    const pathLen = path.length;
+    const patternLen = pattern.length;
+
+    // 快速失败: 长度不匹配(除非以 ** 结尾)
+    if (pathLen !== patternLen && (patternLen === 0 || pattern[patternLen - 1] !== '**')) {
         return false;
     }
-    let fPattern = [...pattern]
-    if(pattern.length >0 && pattern[pattern.length-1] === '**'){
-        fPattern.splice(pattern.length-1,1,...Array.from<string>({
-            length: path.length-pattern.length+1
-        }).fill('*'))
-    }
-    for(let i=0;i<path.length;i++){
-        if(fPattern[i]==='*'){
-            continue
+
+    // 处理 ** 通配符: 直接检查前缀匹配
+    if (patternLen > 0 && pattern[patternLen - 1] === '**') {
+        // 检查前 n-1 个元素是否匹配
+        for (let i = 0; i < patternLen - 1; i++) {
+            if (pattern[i] !== '*' && pattern[i] !== path[i]) {
+                return false;
+            }
         }
-        if(fPattern[i]!==path[i]){
-            return false
+        return true;
+    }
+
+    // 普通匹配: 逐元素比较
+    for (let i = 0; i < pathLen; i++) {
+        if (pattern[i] !== '*' && pattern[i] !== path[i]) {
+            return false;
         }
     }
-    return true
+
+    return true;
 }
 
- 
+// export function isPathMatched(path:string[],pattern:string[]):boolean{
+//     if(path.length !== pattern.length && (path.length>0 && pattern[pattern.length-1]!=='**') ){
+//         return false;
+//     }
+//     let fPattern = [...pattern]
+//     if(pattern.length >0 && pattern[pattern.length-1] === '**'){
+//         fPattern.splice(pattern.length-1,1,...Array.from<string>({
+//             length: path.length-pattern.length+1
+//         }).fill('*'))
+//     }
+//     for(let i=0;i<path.length;i++){
+//         if(fPattern[i]==='*'){
+//             continue
+//         }
+//         if(fPattern[i]!==path[i]){
+//             return false
+//         }
+//     }
+//     return true
+// }

packages/native/src/utils/removeItem.ts

@@ -1,16 +1,31 @@
-
-
-export function removeItem(arr:any[],condition:(item:any)=>boolean){
-    let index:number[] = []
-    while (true) {
-        const i = arr.findIndex((item)=>{            
-            return condition(item)
-        })
-        if(i === -1) {
-            index.push(i)
-            break
-        }        
-        arr.splice(i,1)
+/**
+ * 从数组中移除满足条件的元素并返回被移除元素的索引
+ * @param {any[]} arr - 需要操作的数组
+ * @param {(item: any) => boolean} condition - 判断元素是否应该被移除的条件函数
+ * @returns {number[]} 被移除元素在原数组中的索引集合
+ */
+export function removeItem(arr: any[], condition: (item: any) => boolean) {
+    const removedIndices: number[] = [];
+    // 从后往前遍历，避免删除元素时索引变化的问题
+    for (let i = arr.length - 1; i >= 0; i--) {
+        if (condition(arr[i])) {
+            removedIndices.push(i); // 记录原始索引
+            arr.splice(i, 1); // 删除元素
+        }
     }
-    return index
-}
+
+    // 因为是从后往前遍历的，所以索引是降序的，需要反转得到升序结果
+    return removedIndices.reverse();
+}
+// let index: number[] = [];
+// while (true) {
+//     const i = arr.findIndex((item) => {
+//         return condition(item);
+//     });
+//     if (i === -1) {
+//         index.push(i);
+//         break;
+//     }
+//     arr.splice(i, 1);
+// }
+// return index;