leetcode1583. Count Unhappy Friends
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题目如下:
You are given a list of
preferences
forn
friends, wheren
is always even.For each person
i
,preferences[i]
contains a list of friends sorted in the order of preference. In other words, a friend earlier in the list is more preferred than a friend later in the list. Friends in each list are denoted by integers from0
ton-1
.All the friends are divided into pairs. The pairings are given in a list
pairs
, wherepairs[i] = [xi, yi]
denotesxi
is paired withyi
andyi
is paired withxi
.However, this pairing may cause some of the friends to be unhappy. A friend
x
is unhappy ifx
is paired withy
and there exists a friendu
who is paired withv
but:
x
prefersu
overy
, andu
prefersx
overv
.Return the number of unhappy friends.
Example 1:
Input: n = 4, preferences = [[1, 2, 3], [3, 2, 0], [3, 1, 0], [1, 2, 0]], pairs = [[0, 1], [2, 3]] Output: 2 Explanation: Friend 1 is unhappy because: - 1 is paired with 0 but prefers 3 over 0, and - 3 prefers 1 over 2. Friend 3 is unhappy because: - 3 is paired with 2 but prefers 1 over 2, and - 1 prefers 3 over 0. Friends 0 and 2 are happy.Example 2:
Input: n = 2, preferences = [[1], [0]], pairs = [[1, 0]] Output: 0 Explanation: Both friends 0 and 1 are happy.Example 3:
Input: n = 4, preferences = [[1, 3, 2], [2, 3, 0], [1, 3, 0], [0, 2, 1]], pairs = [[1, 3], [0, 2]] Output: 4Constraints:
2 <= n <= 500
n
is even.preferences.length == n
preferences[i].length == n - 1
0 <= preferences[i][j] <= n - 1
preferences[i]
does not containi
.- All values in
preferences[i]
are unique.pairs.length == n/2
pairs[i].length == 2
xi != yi
0 <= xi, yi <= n - 1
- Each person is contained in exactly one pair.
解题思路:本题很简单,只要为每个朋友分配一个权值即可,grid_preferences[i][j] = v 表示对于i来说,朋友j的权值是v,v值越小越亲近。
代码如下:
class Solution(object): def unhappyFriends(self, n, preferences, pairs): """ :type n: int :type preferences: List[List[int]] :type pairs: List[List[int]] :rtype: int """ dic_unHappy = {} grid_preferences = [[0] * n for _ in range(n)] for i in range(len(preferences)): for j in range(len(preferences[i])): grid_preferences[i][preferences[i][j]] = j for i in range(len(pairs)): for j in range(len(pairs)): if i == j:continue x = pairs[i][0] y = pairs[i][1] u = pairs[j][0] v = pairs[j][1] \'\'\' (x,y), (u,v) x is unhappy if x is paired with y and there exists a friend u who is paired with v but: x prefers u over y, and u prefers x over v. \'\'\' if grid_preferences[x][u] < grid_preferences[x][y] and grid_preferences[u][x] < grid_preferences[u][v]: dic_unHappy[x] = 1 if grid_preferences[x][v] < grid_preferences[x][y] and grid_preferences[v][x] < grid_preferences[v][u]: dic_unHappy[x] = 1 if grid_preferences[y][u] < grid_preferences[y][x] and grid_preferences[u][y] < grid_preferences[u][v]: dic_unHappy[y] = 1 if grid_preferences[y][v] < grid_preferences[y][x] and grid_preferences[v][y] < grid_preferences[v][u]: dic_unHappy[y] = 1 return len(dic_unHappy)
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