Question

473. Matchsticks to Square

中文文档

Description

You are given an integer array matchsticks where matchsticks[i] is the length of the ith matchstick. You want to use all the matchsticks to make one square. You should not break any stick, but you can link them up, and each matchstick must be used exactly one time.

Return true if you can make this square and false otherwise.

 

Example 1:

Input: matchsticks = [1,1,2,2,2]
Output: true
Explanation: You can form a square with length 2, one side of the square came two sticks with length 1.

Example 2:

Input: matchsticks = [3,3,3,3,4]
Output: false
Explanation: You cannot find a way to form a square with all the matchsticks.

 

Constraints:

• 1 <= matchsticks.length <= 15
• 1 <= matchsticks[i] <= 108

Solutions

Solution 1

class Solution:
  def makesquare(self, matchsticks: List[int]) -> bool:
    def dfs(u):
      if u == len(matchsticks):
        return True
      for i in range(4):
        if i > 0 and edges[i - 1] == edges[i]:
          continue
        edges[i] += matchsticks[u]
        if edges[i] <= x and dfs(u + 1):
          return True
        edges[i] -= matchsticks[u]
      return False

    x, mod = divmod(sum(matchsticks), 4)
    if mod or x < max(matchsticks):
      return False
    edges = [0] * 4
    matchsticks.sort(reverse=True)
    return dfs(0)

class Solution {
  public boolean makesquare(int[] matchsticks) {
    int s = 0, mx = 0;
    for (int v : matchsticks) {
      s += v;
      mx = Math.max(mx, v);
    }
    int x = s / 4, mod = s % 4;
    if (mod != 0 || x < mx) {
      return false;
    }
    Arrays.sort(matchsticks);
    int[] edges = new int[4];
    return dfs(matchsticks.length - 1, x, matchsticks, edges);
  }

  private boolean dfs(int u, int x, int[] matchsticks, int[] edges) {
    if (u < 0) {
      return true;
    }
    for (int i = 0; i < 4; ++i) {
      if (i > 0 && edges[i - 1] == edges[i]) {
        continue;
      }
      edges[i] += matchsticks[u];
      if (edges[i] <= x && dfs(u - 1, x, matchsticks, edges)) {
        return true;
      }
      edges[i] -= matchsticks[u];
    }
    return false;
  }
}

class Solution {
public:
  bool makesquare(vector<int>& matchsticks) {
    int s = 0, mx = 0;
    for (int& v : matchsticks) {
      s += v;
      mx = max(mx, v);
    }
    int x = s / 4, mod = s % 4;
    if (mod != 0 || x < mx) return false;
    sort(matchsticks.begin(), matchsticks.end(), greater<int>());
    vector<int> edges(4);
    return dfs(0, x, matchsticks, edges);
  }

  bool dfs(int u, int x, vector<int>& matchsticks, vector<int>& edges) {
    if (u == matchsticks.size()) return true;
    for (int i = 0; i < 4; ++i) {
      if (i > 0 && edges[i - 1] == edges[i]) continue;
      edges[i] += matchsticks[u];
      if (edges[i] <= x && dfs(u + 1, x, matchsticks, edges)) return true;
      edges[i] -= matchsticks[u];
    }
    return false;
  }
};

func makesquare(matchsticks []int) bool {
  s := 0
  for _, v := range matchsticks {
    s += v
  }
  if s%4 != 0 {
    return false
  }
  sort.Sort(sort.Reverse(sort.IntSlice(matchsticks)))
  edges := make([]int, 4)
  var dfs func(u, x int) bool
  dfs = func(u, x int) bool {
    if u == len(matchsticks) {
      return true
    }
    for i := 0; i < 4; i++ {
      if i > 0 && edges[i-1] == edges[i] {
        continue
      }
      edges[i] += matchsticks[u]
      if edges[i] <= x && dfs(u+1, x) {
        return true
      }
      edges[i] -= matchsticks[u]
    }
    return false
  }
  return dfs(0, s/4)
}

impl Solution {
  pub fn makesquare(matchsticks: Vec<i32>) -> bool {
    let mut matchsticks = matchsticks;

    fn dfs(matchsticks: &Vec<i32>, edges: &mut [i32; 4], u: usize, x: i32) -> bool {
      if u == matchsticks.len() {
        return true;
      }
      for i in 0..4 {
        if i > 0 && edges[i - 1] == edges[i] {
          continue;
        }
        edges[i] += matchsticks[u];
        if edges[i] <= x && dfs(matchsticks, edges, u + 1, x) {
          return true;
        }
        edges[i] -= matchsticks[u];
      }
      false
    }

    let sum: i32 = matchsticks.iter().sum();
    if sum % 4 != 0 {
      return false;
    }
    matchsticks.sort_by(|x, y| y.cmp(x));
    let mut edges = [0; 4];

    dfs(&matchsticks, &mut edges, 0, sum / 4)
  }
}

Solution 2

class Solution:
  def makesquare(self, matchsticks: List[int]) -> bool:
    @cache
    def dfs(state, t):
      if state == (1 << len(matchsticks)) - 1:
        return True
      for i, v in enumerate(matchsticks):
        if state & (1 << i):
          continue
        if t + v > s:
          break
        if dfs(state | (1 << i), (t + v) % s):
          return True
      return False

    s, mod = divmod(sum(matchsticks), 4)
    matchsticks.sort()
    if mod:
      return False
    return dfs(0, 0)