82 lines
2.6 KiB
C++
82 lines
2.6 KiB
C++
// Source : https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/
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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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* Design an algorithm to find the maximum profit. You may complete as many transactions as you like
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* (i.e., buy one and sell one share of the stock multiple times).
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*
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* Note: You may not engage in multiple transactions at the same time (i.e., you must sell the stock
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* before you buy again).
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*
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* Example 1:
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*
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* Input: [7,1,5,3,6,4]
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* Output: 7
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* Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
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* Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
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*
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* Example 2:
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*
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* Input: [1,2,3,4,5]
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* Output: 4
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* Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
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* Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
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* engaging multiple transactions at the same time. You must sell before buying again.
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*
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* Example 3:
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*
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* Input: [7,6,4,3,1]
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* Output: 0
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* Explanation: In this case, no transaction is done, i.e. max profit = 0.
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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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return maxProfit02(prices);
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return maxProfit01(prices);
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}
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// Solution 1
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// find all of ranges: which start a valley with the nearest peak after
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// add their delta together
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//
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int maxProfit01(vector<int> &prices) {
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int max = 0;
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int low = -1;
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int len = prices.size();
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for (int i=0; i < len - 1; i++){
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//meet the valley, then goes up
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if (prices[i] < prices[i+1] && low < 0 ) {
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low = i;
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}
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//meet the peak, then goes down
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if (prices[i] > prices[i+1] && low >= 0) {
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max += ( prices[i] - prices[low] ) ;
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low = -1; // reset the `low`
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}
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}
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// edge case
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if ( low >= 0 ) {
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max += ( prices[prices.size()-1] - prices[low] );
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}
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return max;
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}
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// Solution 2
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// if we find we can earn money, we just sell
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int maxProfit02(vector<int>& prices) {
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int profit = 0 ;
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for(int i=1; i< prices.size(); i++) {
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profit += max(0, prices[i] - prices[i-1]);
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}
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return profit;
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}
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};
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