1475.Final Prices With a Special Discount in a Shop

Tags: Easy Stack

Links: https://leetcode.com/problems/final-prices-with-a-special-discount-in-a-shop/

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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]

Constraints:

• 1 <= prices.length <= 500
• 1 <= prices[i] <= 10^3

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class Solution {
public:
    vector<int> finalPrices(vector<int>& prices) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int n = prices.size();
        vector<int> res(n, 0);
        prices.push_back(0); //保证所有数据弹出
        stack<int> s;
        for (int i = 0; i <= n; ++i) {
            if (s.empty()) {
                s.push(i);
            }
            else {
                while (!s.empty() && prices[s.top()] >= prices[i]) {
                    int pos = s.top(); s.pop();
                    res[pos] = prices[pos] - prices[i];
                }
                s.push(i);
            }
        }

        return res;
    }
};

单调栈的板子题，时间复杂度$O(n)$。

从当前位置开始，找第一个小于当前值的数，所以栈内下标对应的元素从栈顶到栈底是单调递减的。