69 lines
2.0 KiB
C++
69 lines
2.0 KiB
C++
// Source : https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock/
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// Author : Hao Chen
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// Date : 2014-06-18
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/**********************************************************************************
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*
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* Say you have an array for which the ith element is the price of a given stock on day i.
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*
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* If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock),
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* design an algorithm to find the maximum profit.
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*
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**********************************************************************************/
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class Solution {
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public:
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//
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// This solution is O(1) space dynamic programming
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//
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// We can make sure the max profit at least be ZERO. So,
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// 1) we have two pointers (begin & end )
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// 2) if prices[end] - prices[begin] > 0, then move the "end" pointer to next
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// 3) if prices[end] - prices[begin] <= 0, then move the "begin" pointer to current posstion.
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// 4) tracking the max profit
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//
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// Notes:
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// Some people think find the highest-price & lowest-price, this is wrong.
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// Because the highest-price must be after lowest-price
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//
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int maxProfit(vector<int> &prices) {
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int max=0, begin=0, end=0, delta=0;
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for (int i=0; i<prices.size(); i++) {
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end = i;
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delta = prices[end] - prices[begin];
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if (delta <= 0){
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begin = i;
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}
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if ( delta > max ){
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max = delta;
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}
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}
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return max;
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}
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};
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class Solution {
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public:
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int maxProfit(vector<int>& prices) {
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int buy = INT_MAX;
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int profit = 0;
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for (auto p : prices) {
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// Keep tracking the previous lowest price
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buy = min (buy, p);
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// Keep tacking the current max profit
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profit = max(profit, p - buy);
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}
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return profit;
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}
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};
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