找到最近的Boss算法题

最新推荐文章于 2025-01-14 13:54:11 发布

dingyanpeng

最新推荐文章于 2025-01-14 13:54:11 发布

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分类专栏： iOS 文章标签：找到最近的Boss

本文链接：https://blog.youkuaiyun.com/dingyanpeng/article/details/50331745

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该博客介绍了一个算法问题，即如何在层级关系中找到两个人最近的共同上级。通过递归方法，从顶层开始向下搜索，判断当前管理者是否包含任一成员或两者都在同一分支上，最终找到最近的共同上级。

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//层级关系，任意给两个元素，找到最近的、公共的上一级

boss

membera1 members membera3

memberb1 memberb2

#import "ViewController.h"

@interface ViewController ()

@property (nonatomic,strong)NSArray *Bill;

@property (nonatomic,strong)NSArray *Dom;

@property (nonatomic,strong)NSArray *Samir;

@property (nonatomic,strong)NSArray *Michael;

@property (nonatomic,strong)NSArray *Bob;

@property (nonatomic,strong)NSArray *Peter;

@property (nonatomic,strong)NSArray *Porter;

@property (nonatomic,strong) NSArray *Flisa;

@property (nonatomic,strong)NSArray *Milton;

@property (nonatomic,strong)NSArray *Nina;

@end

@implementation ViewController

- (void)viewDidLoad {

[superviewDidLoad];

//每个数组代表一个人，数组内包含它的下属，

#pragma mark: 注意：数组的第一个元素是自己的名字，遍历数组时，从 i=1开始

//最底层

self.Nina =@[@"Nina"];

self.Milton =@[@"Milton"];

//上一层

self.Bob =@[@"Bob"];

self.Peter =@[@"Peter",self.Nina,self.Milton];

self.Porter =@[@"Porter"];

self.Flisa =@[@"Flisa"];

//上一层

self.Dom =@[@"Dom",self.Bob,self.Peter,self.Porter];

self.Samir =@[@"Samir"];

self.Michael =@[@"Michael",self.Flisa];

//最顶层

self.Bill =@[@"Bill",self.Dom,self.Samir,self.Michael];

//从顶层开始往下检索，第一次调用这个方法时，HigerManager参数必须用 self.Bill

NSArray *closest = [selfclosestCommonManagerByMember1:self.NinaMember2:self.MichaelHigherManager:self.Bill];

//打印名字，看结果是否正确

NSLog(@"%@", closest.firstObject);

}

#pragma mark:执行这个方法

- (NSArray *)closestCommonManagerByMember1:(NSArray *)member1 Member2:(NSArray *)member2 HigherManager:(NSArray *)manager{

BOOL contains1 = [manager containsObject:member1];

BOOL contains2 = [manager containsObject:member2];

BOOL special1 = [selfmanager:member1containsMember:member2];

BOOL special2 = [selfmanager:member2containsMember:member1];

if (special1) {

return member1;

}

if (special2) {

return member2;

}

//manager至少直接包含其中一个(只是“父子”关系)，两个member不存在包含关系

if (contains1 || contains2) {

return manager;

}

//manager两个都不直接包含

for (int i=1; i<manager.count; i++) {

BOOL containsForPath1 = [selfmanager:manager[i]containsMember:member1];

BOOL containsForPath2 = [selfmanager:manager[i]containsMember:member2];

//两个member都不在这条分支上

if (!containsForPath1 && !containsForPath2) {

continue;

}

//两个member都在这条分支上

if (containsForPath1 && containsForPath2) {

return [selfclosestCommonManagerByMember1:member1Member2:member2HigherManager:manager[i]];

}

//只有一个member在这条分支上

return manager;

}

return nil;

}

//判断 manager当中是否包含 member,（member可以是manager的儿子，也可以是孙子，也可以是重孙子.....）

- (BOOL)manager:(NSArray *)manager containsMember:(NSArray *)member {

if (manager.count <=1) {

return false;

}

if ([manager containsObject:member]) {

return true;

}

BOOL contains = false;

for (int i=1; i<manager.count; i++) {

contains = contains || [self manager:manager[i] containsMember:member];

}

return contains;

}

@end