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Add solution 1600, change dir 0167、0303、0304、0307、0653、1017
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41
leetcode/1600.Throne-Inheritance/1600. Throne Inheritance.go
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41
leetcode/1600.Throne-Inheritance/1600. Throne Inheritance.go
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package leetcode
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type ThroneInheritance struct {
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king string
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edges map[string][]string
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dead map[string]bool
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}
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func Constructor(kingName string) (t ThroneInheritance) {
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return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}}
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}
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func (t *ThroneInheritance) Birth(parentName, childName string) {
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t.edges[parentName] = append(t.edges[parentName], childName)
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}
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func (t *ThroneInheritance) Death(name string) {
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t.dead[name] = true
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}
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func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
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var preorder func(string)
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preorder = func(name string) {
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if !t.dead[name] {
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res = append(res, name)
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}
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for _, childName := range t.edges[name] {
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preorder(childName)
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}
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}
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preorder(t.king)
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return
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}
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/**
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* Your ThroneInheritance object will be instantiated and called as such:
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* obj := Constructor(kingName);
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* obj.Birth(parentName,childName);
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* obj.Death(name);
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* param_3 := obj.GetInheritanceOrder();
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*/
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package leetcode
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import (
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"fmt"
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"testing"
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)
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func Test_Problem1600(t *testing.T) {
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obj := Constructor("king")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("king", "andy")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("king", "bob")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("king", "catherine")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("andy", "matthew")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("bob", "alex")
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fmt.Printf("obj = %v\n", obj)
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obj.Birth("bob", "asha")
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fmt.Printf("obj = %v\n", obj)
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param2 := obj.GetInheritanceOrder()
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fmt.Printf("param_2 = %v obj = %v\n", param2, obj)
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obj.Death("bob")
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fmt.Printf("obj = %v\n", obj)
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param2 = obj.GetInheritanceOrder()
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fmt.Printf("param_2 = %v obj = %v\n", param2, obj)
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}
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120
leetcode/1600.Throne-Inheritance/README.md
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120
leetcode/1600.Throne-Inheritance/README.md
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# [1600. Throne Inheritance](https://leetcode.com/problems/throne-inheritance/)
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## 题目
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A kingdom consists of a king, his children, his grandchildren, and so on. Every once in a while, someone in the family dies or a child is born.
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The kingdom has a well-defined order of inheritance that consists of the king as the first member. Let's define the recursive function `Successor(x, curOrder)`, which given a person `x` and the inheritance order so far, returns who should be the next person after `x` in the order of inheritance.
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```
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Successor(x, curOrder):
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if x has no children or all of x's children are in curOrder:
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if x is the king return null
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else return Successor(x's parent, curOrder)
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else return x's oldest child who's not in curOrder
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```
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For example, assume we have a kingdom that consists of the king, his children Alice and Bob (Alice is older than Bob), and finally Alice's son Jack.
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1. In the beginning, `curOrder` will be `["king"]`.
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2. Calling `Successor(king, curOrder)` will return Alice, so we append to `curOrder` to get `["king", "Alice"]`.
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3. Calling `Successor(Alice, curOrder)` will return Jack, so we append to `curOrder` to get `["king", "Alice", "Jack"]`.
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4. Calling `Successor(Jack, curOrder)` will return Bob, so we append to `curOrder` to get `["king", "Alice", "Jack", "Bob"]`.
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5. Calling `Successor(Bob, curOrder)` will return `null`. Thus the order of inheritance will be `["king", "Alice", "Jack", "Bob"]`.
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Using the above function, we can always obtain a unique order of inheritance.
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Implement the `ThroneInheritance` class:
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- `ThroneInheritance(string kingName)` Initializes an object of the `ThroneInheritance` class. The name of the king is given as part of the constructor.
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- `void birth(string parentName, string childName)` Indicates that `parentName` gave birth to `childName`.
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- `void death(string name)` Indicates the death of `name`. The death of the person doesn't affect the `Successor` function nor the current inheritance order. You can treat it as just marking the person as dead.
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- `string[] getInheritanceOrder()` Returns a list representing the current order of inheritance **excluding** dead people.
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**Example 1:**
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```
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Input
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["ThroneInheritance", "birth", "birth", "birth", "birth", "birth", "birth", "getInheritanceOrder", "death", "getInheritanceOrder"]
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[["king"], ["king", "andy"], ["king", "bob"], ["king", "catherine"], ["andy", "matthew"], ["bob", "alex"], ["bob", "asha"], [null], ["bob"], [null]]
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Output
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[null, null, null, null, null, null, null, ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"], null, ["king", "andy", "matthew", "alex", "asha", "catherine"]]
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Explanation
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ThroneInheritance t= new ThroneInheritance("king"); // order:king
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t.birth("king", "andy"); // order: king >andy
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t.birth("king", "bob"); // order: king > andy >bob
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t.birth("king", "catherine"); // order: king > andy > bob >catherine
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t.birth("andy", "matthew"); // order: king > andy >matthew > bob > catherine
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t.birth("bob", "alex"); // order: king > andy > matthew > bob >alex > catherine
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t.birth("bob", "asha"); // order: king > andy > matthew > bob > alex >asha > catherine
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t.getInheritanceOrder(); // return ["king", "andy", "matthew", "bob", "alex", "asha", "catherine"]
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t.death("bob"); // order: king > andy > matthew >bob > alex > asha > catherine
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t.getInheritanceOrder(); // return ["king", "andy", "matthew", "alex", "asha", "catherine"]
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```
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**Constraints:**
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- `1 <= kingName.length, parentName.length, childName.length, name.length <= 15`
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- `kingName`, `parentName`, `childName`, and `name` consist of lowercase English letters only.
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- All arguments `childName` and `kingName` are **distinct**.
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- All `name` arguments of `death` will be passed to either the constructor or as `childName` to `birth` first.
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- For each call to `birth(parentName, childName)`, it is guaranteed that `parentName` is alive.
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- At most `105` calls will be made to `birth` and `death`.
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- At most `10` calls will be made to `getInheritanceOrder`.
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## 题目大意
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一个王国里住着国王、他的孩子们、他的孙子们等等。每一个时间点,这个家庭里有人出生也有人死亡。这个王国有一个明确规定的皇位继承顺序,第一继承人总是国王自己。我们定义递归函数 Successor(x, curOrder) ,给定一个人 x 和当前的继承顺序,该函数返回 x 的下一继承人。
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## 解题思路
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- 这道题思路不难。先将国王每个孩子按照顺序存在一个 map 中,然后每个国王的孩子还存在父子关系,同理也按顺序存在 map 中。执行 GetInheritanceOrder() 函数时,将国王的孩子按顺序遍历,如果每个孩子还有孩子,递归遍历到底。如果把继承关系看成一棵树,此题便是多叉树的先根遍历的问题。
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## 代码
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```go
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package leetcode
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type ThroneInheritance struct {
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king string
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edges map[string][]string
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dead map[string]bool
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}
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func Constructor(kingName string) (t ThroneInheritance) {
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return ThroneInheritance{kingName, map[string][]string{}, map[string]bool{}}
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}
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func (t *ThroneInheritance) Birth(parentName, childName string) {
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t.edges[parentName] = append(t.edges[parentName], childName)
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}
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func (t *ThroneInheritance) Death(name string) {
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t.dead[name] = true
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}
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func (t *ThroneInheritance) GetInheritanceOrder() (res []string) {
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var preorder func(string)
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preorder = func(name string) {
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if !t.dead[name] {
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res = append(res, name)
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}
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for _, childName := range t.edges[name] {
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preorder(childName)
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}
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}
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preorder(t.king)
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return
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}
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/**
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* Your ThroneInheritance object will be instantiated and called as such:
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* obj := Constructor(kingName);
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* obj.Birth(parentName,childName);
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* obj.Death(name);
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* param_3 := obj.GetInheritanceOrder();
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*/
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```
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