数组删除操作
总结一下
- (void)viewDidLoad { [super viewDidLoad]; self.testArray = [NSMutableArray arrayWithObjects:@"cat",@"dog",@"man",@"dog",@"dog", nil]; } - (void)touchesBegan:(NSSet<UITouch *> *)touches withEvent:(UIEvent *)event{ // [self removeObjectWithInset]; // [self removeObjWithIndex]; // [self removeObjWithObj]; // [self removeObj]; // [self removeTheObjWithInset]; NSLog(@"%@",self.testArray); } /** 通过index删除 */ - (void)removeObjWithIndex{ for (int i = 0; i< self.testArray.count; i++) { NSString *var = self.testArray[i]; if ([var isEqualToString:@"dog"]) { [self.testArray removeObjectAtIndex:i]; //如果删除所有isEqual的obj 不加Break //如果删除第一个则加上break break; } } } /** 不加break会造成crash 即使加上break也会删除所有的dog */ - (void)removeObjWithObj{ for (NSString *str in self.testArray) { if ([str isEqualToString:@"dog"]) { [self.testArray removeObject:str]; break; } } } /** 这样直接删除也是可以的 */ - (void)removeObj{ [self.testArray removeObject:@"dog"]; } /** 通过NSInset删除 */ - (void)removeObjectWithInset{ NSIndexSet *indexSet = [self.testArray indexesOfObjectsPassingTest:^BOOL(id _Nonnull obj, NSUInteger idx, BOOL * _Nonnull stop) { return [obj isEqualToString:@"dog"]; }]; [self.testArray removeObjectsAtIndexes:indexSet]; } /** 删除第二个dog */ - (void)removeTheObjWithInset{ NSMutableArray *idxArray = [NSMutableArray array]; for (int i = 0; i< self.testArray.count; i++) { NSString *str = self.testArray[i]; if ([str isEqualToString:@"dog"]) { [idxArray addObject: [NSString stringWithFormat:@"%d",i]]; } } NSInteger getIndex; getIndex = [idxArray[1] integerValue]; [self.testArray removeObjectAtIndex:getIndex]; }