描述
Given the array prices where prices[i] is the price of the ith item in a shop. There is a special discount for items in the shop, if you buy the ith item, then you will receive a discount equivalent to prices[j] where j is the minimum index such that j > i and prices[j] <= prices[i], otherwise, you will not receive any discount at all.
Return an array where the ith element is the final price you will pay for the ith item of the shop considering the special discount.
Example 1:
Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Explanation:
For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4.
For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2.
For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4.
For items 3 and 4 you will not receive any discount at all.
Example 2:
Input: prices = [1,2,3,4,5]
Output: [1,2,3,4,5]
Explanation: In this case, for all items, you will not receive any discount at all.
Example 3:
Input: prices = [10,1,1,6]
Output: [9,0,1,6]
Note:
1 <= prices.length <= 500
1 <= prices[i] <= 10^3
解析
根据题意，物品 i 价值为 prices[i] ，找出最小的索引 j 使得 j>i ，且 prices[j]<prices[i] ，如果有这个元素则将其值作为可以享受的折扣，最后则将索引为 i 的物品价值改为 prices[i]-prices[j] 即可，遍历每个元素都如此操作，最后返回结果。
解答
class Solution(object):
def finalPrices(self, prices):
"""
:type prices: List[int]
:rtype: List[int]
"""
n = len(prices)
for i in range(n-1):
j=i+1
while j<n and prices[j]>prices[i]:
j += 1
if j<n:
prices[i] -= prices[j]
return prices
运行结果
Runtime: 40 ms, faster than 77.21% of Python online submissions for Final Prices With a Special Discount in a Shop.
Memory Usage: 13.4 MB, less than 75.89% of Python online submissions for Final Prices With a Special Discount in a Shop.
原题链接：https://leetcode.com/problems/final-prices-with-a-special-discount-in-a-shop/