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Commit d52913b

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Merge pull request #383 from 0xff-dev/834
Add solution and test-cases for problem 834
2 parents 66537d2 + a7017f4 commit d52913b

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‎leetcode/801-900/0834.Sum-of-Distances-in-Tree/README.md‎

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# [834.Sum of Distances in Tree][title]
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> [!WARNING|style:flat]
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> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
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## Description
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There is an undirected connected tree with n nodes labeled from `0` to `n - 1` and `n - 1` edges.
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You are given the integer `n` and the array `edges` where `edges[i] = [ai, bi]` indicates that there is an edge between nodes a<sub>i</sub> and b<sub>i</sub> in the tree.
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Return an array `answer` of length `n` where `answer[i]` is the sum of the distances between the i<sup>th</sup> node in the tree and all other nodes.
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**Example 1:**
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**Example 1:**
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![example1](./lc-sumdist1.jpg)
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```
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Input: a = "11", b = "1"
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Output: "100"
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Input: n = 6, edges = [[0,1],[0,2],[2,3],[2,4],[2,5]]
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Output: [8,12,6,10,10,10]
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Explanation: The tree is shown above.
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We can see that dist(0,1) + dist(0,2) + dist(0,3) + dist(0,4) + dist(0,5)
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equals 1 +たす 1 +たす 2 +たす 2 +たす 2 = 8.
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Hence, answer[0] = 8, and so on.
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```
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## 题意
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> ...
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**Example 2:**
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## 题解
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![example2](./lc-sumdist2.jpg)
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### 思路1
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> ...
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Sum of Distances in Tree
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```go
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```
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Input: n = 1, edges = []
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Output: [0]
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```
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**Example 3:**
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![example3](./lc-sumdist3.jpg)
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```
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Input: n = 2, edges = [[1,0]]
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Output: [1,1]
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```
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## 结语
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package Solution
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func Solution(x bool) bool {
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return x
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// 借鉴官方算法
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func Solution(n int, edges [][]int) []int {
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adj := make(map[int]map[int]struct{})
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for u := 0; u < n; u++ {
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adj[u] = make(map[int]struct{})
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}
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for _, e := range edges {
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adj[e[0]][e[1]] = struct{}{}
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adj[e[1]][e[0]] = struct{}{}
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}
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ans := make([]int, n)
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count := make([]int, n)
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for i := 0; i < n; i++ {
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count[i] = 1
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}
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var dfs func(int, int)
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var dfs1 func(int, int)
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dfs = func(node, parent int) {
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for child := range adj[node] {
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if child != parent {
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dfs(child, node)
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count[node] += count[child]
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ans[node] += ans[child] + count[child]
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}
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}
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}
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dfs1 = func(node, parent int) {
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for child := range adj[node] {
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if child != parent {
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ans[child] = ans[node] - count[child] + n - count[child]
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dfs1(child, node)
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}
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}
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}
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dfs(0, -1)
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dfs1(0, -1)
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return ans
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}

‎leetcode/801-900/0834.Sum-of-Distances-in-Tree/Solution_test.go‎

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// 测试用例
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cases := []struct {
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name string
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inputs bool
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expect bool
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n int
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edges [][]int
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expect []int
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}{
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{"TestCase", true, true},
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{"TestCase", true, true},
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{"TestCase", false, false},
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{"TestCase1", 6, [][]int{{0, 1}, {0, 2}, {2, 3}, {2, 4}, {2, 5}}, []int{8, 12, 6, 10, 10, 10}},
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{"TestCase2", 1, [][]int{}, []int{0}},
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{"TestCase3", 2, [][]int{{1, 0}}, []int{1, 1}},
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{"TestCase4", 4, [][]int{{2, 0}, {3, 1}, {2, 1}}, []int{6, 4, 4, 6}},
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{"TestCase5", 5, [][]int{{0, 4}, {1, 3}, {1, 2}, {0, 2}}, []int{7, 7, 6, 10, 10}},
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}
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// 开始测试
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for i, c := range cases {
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t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
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got := Solution(c.inputs)
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got := Solution(c.n, c.edges)
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if !reflect.DeepEqual(got, c.expect) {
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t.Fatalf("expected: %v, but got: %v, with inputs: %v",
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c.expect, got, c.inputs)
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t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
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c.expect, got, c.n, c.edges)
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}
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})
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}
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}
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//压力测试
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//压力测试
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func BenchmarkSolution(b *testing.B) {
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}
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//使用案列
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//使用案列
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func ExampleSolution() {
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}
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