Question

1743. Restore the Array From Adjacent Pairs

中文文档

Description

There is an integer array nums that consists of n unique elements, but you have forgotten it. However, you do remember every pair of adjacent elements in nums.

You are given a 2D integer array adjacentPairs of size n - 1 where each adjacentPairs[i] = [ui, vi] indicates that the elements ui and vi are adjacent in nums.

It is guaranteed that every adjacent pair of elements nums[i] and nums[i+1] will exist in adjacentPairs, either as [nums[i], nums[i+1]] or [nums[i+1], nums[i]]. The pairs can appear in any order.

Return _the original array _nums_. If there are multiple solutions, return any of them_.

 

Example 1:

Input: adjacentPairs = [[2,1],[3,4],[3,2]]
Output: [1,2,3,4]
Explanation: This array has all its adjacent pairs in adjacentPairs.
Notice that adjacentPairs[i] may not be in left-to-right order.

Example 2:

Input: adjacentPairs = [[4,-2],[1,4],[-3,1]]
Output: [-2,4,1,-3]
Explanation: There can be negative numbers.
Another solution is [-3,1,4,-2], which would also be accepted.

Example 3:

Input: adjacentPairs = [[100000,-100000]]
Output: [100000,-100000]

 

Constraints:

• nums.length == n
• adjacentPairs.length == n - 1
• adjacentPairs[i].length == 2
• 2 <= n <= 105
• -105 <= nums[i], ui, vi <= 105
• There exists some nums that has adjacentPairs as its pairs.

Solutions

Solution 1

class Solution:
  def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
    g = defaultdict(list)
    for a, b in adjacentPairs:
      g[a].append(b)
      g[b].append(a)
    n = len(adjacentPairs) + 1
    ans = [0] * n
    for i, v in g.items():
      if len(v) == 1:
        ans[0] = i
        ans[1] = v[0]
        break
    for i in range(2, n):
      v = g[ans[i - 1]]
      ans[i] = v[0] if v[1] == ans[i - 2] else v[1]
    return ans

class Solution {
  public int[] restoreArray(int[][] adjacentPairs) {
    int n = adjacentPairs.length + 1;
    Map<Integer, List<Integer>> g = new HashMap<>();
    for (int[] e : adjacentPairs) {
      int a = e[0], b = e[1];
      g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
      g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
    }
    int[] ans = new int[n];
    for (Map.Entry<Integer, List<Integer>> entry : g.entrySet()) {
      if (entry.getValue().size() == 1) {
        ans[0] = entry.getKey();
        ans[1] = entry.getValue().get(0);
        break;
      }
    }
    for (int i = 2; i < n; ++i) {
      List<Integer> v = g.get(ans[i - 1]);
      ans[i] = v.get(1) == ans[i - 2] ? v.get(0) : v.get(1);
    }
    return ans;
  }
}

class Solution {
public:
  vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
    int n = adjacentPairs.size() + 1;
    unordered_map<int, vector<int>> g;
    for (auto& e : adjacentPairs) {
      int a = e[0], b = e[1];
      g[a].push_back(b);
      g[b].push_back(a);
    }
    vector<int> ans(n);
    for (auto& [k, v] : g) {
      if (v.size() == 1) {
        ans[0] = k;
        ans[1] = v[0];
        break;
      }
    }
    for (int i = 2; i < n; ++i) {
      auto v = g[ans[i - 1]];
      ans[i] = v[0] == ans[i - 2] ? v[1] : v[0];
    }
    return ans;
  }
};

func restoreArray(adjacentPairs [][]int) []int {
  n := len(adjacentPairs) + 1
  g := map[int][]int{}
  for _, e := range adjacentPairs {
    a, b := e[0], e[1]
    g[a] = append(g[a], b)
    g[b] = append(g[b], a)
  }
  ans := make([]int, n)
  for k, v := range g {
    if len(v) == 1 {
      ans[0] = k
      ans[1] = v[0]
      break
    }
  }
  for i := 2; i < n; i++ {
    v := g[ans[i-1]]
    ans[i] = v[0]
    if v[0] == ans[i-2] {
      ans[i] = v[1]
    }
  }
  return ans
}

public class Solution {
  public int[] RestoreArray(int[][] adjacentPairs) {
    int n = adjacentPairs.Length + 1;
    Dictionary<int, List<int>> g = new Dictionary<int, List<int>>();

    foreach (int[] e in adjacentPairs) {
      int a = e[0], b = e[1];
      if (!g.ContainsKey(a)) {
        g[a] = new List<int>();
      }
      if (!g.ContainsKey(b)) {
        g[b] = new List<int>();
      }
      g[a].Add(b);
      g[b].Add(a);
    }

    int[] ans = new int[n];

    foreach (var entry in g) {
      if (entry.Value.Count == 1) {
        ans[0] = entry.Key;
        ans[1] = entry.Value[0];
        break;
      }
    }

    for (int i = 2; i < n; ++i) {
      List<int> v = g[ans[i - 1]];
      ans[i] = v[1] == ans[i - 2] ? v[0] : v[1];
    }

    return ans;
  }
}

Solution 2

class Solution:
  def restoreArray(self, adjacentPairs: List[List[int]]) -> List[int]:
    def dfs(i, fa):
      ans.append(i)
      for j in g[i]:
        if j != fa:
          dfs(j, i)

    g = defaultdict(list)
    for a, b in adjacentPairs:
      g[a].append(b)
      g[b].append(a)
    i = next(i for i, v in g.items() if len(v) == 1)
    ans = []
    dfs(i, 1e6)
    return ans

class Solution {
  private Map<Integer, List<Integer>> g = new HashMap<>();
  private int[] ans;

  public int[] restoreArray(int[][] adjacentPairs) {
    for (var e : adjacentPairs) {
      int a = e[0], b = e[1];
      g.computeIfAbsent(a, k -> new ArrayList<>()).add(b);
      g.computeIfAbsent(b, k -> new ArrayList<>()).add(a);
    }
    int n = adjacentPairs.length + 1;
    ans = new int[n];
    for (var e : g.entrySet()) {
      if (e.getValue().size() == 1) {
        dfs(e.getKey(), 1000000, 0);
        break;
      }
    }
    return ans;
  }

  private void dfs(int i, int fa, int k) {
    ans[k++] = i;
    for (int j : g.get(i)) {
      if (j != fa) {
        dfs(j, i, k);
      }
    }
  }
}

class Solution {
public:
  vector<int> restoreArray(vector<vector<int>>& adjacentPairs) {
    unordered_map<int, vector<int>> g;
    for (auto& e : adjacentPairs) {
      int a = e[0], b = e[1];
      g[a].emplace_back(b);
      g[b].emplace_back(a);
    }
    int n = adjacentPairs.size() + 1;
    vector<int> ans;
    function<void(int, int)> dfs = [&](int i, int fa) {
      ans.emplace_back(i);
      for (int& j : g[i]) {
        if (j != fa) {
          dfs(j, i);
        }
      }
    };
    for (auto& [i, v] : g) {
      if (v.size() == 1) {
        dfs(i, 1e6);
        break;
      }
    }
    return ans;
  }
};

func restoreArray(adjacentPairs [][]int) []int {
  g := map[int][]int{}
  for _, e := range adjacentPairs {
    a, b := e[0], e[1]
    g[a] = append(g[a], b)
    g[b] = append(g[b], a)
  }
  ans := []int{}
  var dfs func(i, fa int)
  dfs = func(i, fa int) {
    ans = append(ans, i)
    for _, j := range g[i] {
      if j != fa {
        dfs(j, i)
      }
    }
  }
  for i, v := range g {
    if len(v) == 1 {
      dfs(i, 1000000)
      break
    }
  }
  return ans
}