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135. Candy.py
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135. Candy.py
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# There are N children standing in a line. Each child is assigned a rating value.
# You are giving candies to these children subjected to the following requirements:
# Each child must have at least one candy.
# Children with a higher rating get more candies than their neighbors.
# What is the minimum candies you must give?
# Example 1:
# Input: [1,0,2]
# Output: 5
# Explanation: You can allocate to the first, second and third child with 2, 1, 2 candies respectively.
# Example 2:
# Input: [1,2,2]
# Output: 4
# Explanation: You can allocate to the first, second and third child with 1, 2, 1 candies respectively.
# The third child gets 1 candy because it satisfies the above two conditions.
class Solution:
def candy(self, ratings: List[int]) -> int:
# M1. 蛮力 TLE
n = len(ratings)
candies = [1 for _ in range(n)]
flag = True
res = 0
while flag:
flag = False
for i in range(n):
if i != n-1 and ratings[i] > ratings[i+1] and candies[i] <= candies[i+1]:
candies[i] = candies[i+1] + 1
flag = True
if i > 0 and ratings[i] > ratings[i-1] and candies[i] <= candies[i-1]:
candies[i] = candies[i-1] + 1
flag = True
for candy in candies:
res += candy
return res
# M2. 找规律 双数组
n = len(ratings)
left2right = [1 for _ in range(n)]
right2left = [1 for _ in range(n)]
res = 0
for i in range(1, n):
if ratings[i] > ratings[i-1]:
left2right[i] = left2right[i-1] + 1
for i in range(n-2, -1, -1):
if ratings[i] > ratings[i+1]:
right2left[i] = right2left[i+1] + 1
for i in range(n):
res += left2right[i] if left2right[i] > right2left[i] else right2left[i]
return res
# M3. 找规律 单数组
n = len(ratings)
candies = [1 for _ in range(n)]
for i in range(1, n):
if ratings[i] > ratings[i-1]:
candies[i] = candies[i-1] + 1
for i in range(n-2, -1, -1):
if ratings[i] > ratings[i+1]:
candies[i] = max(candies[i], candies[i+1]+1)
return sum(candies)