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Best Time to Sell and Buy Stocks
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/** | |
* Say you have an array for which the ith element is the price of a given stock on day i. | |
* | |
* If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. | |
*/ | |
public class Solution{ | |
public int maxProfit(int[] prices) { | |
if(prices.length == 0 || prices.length == 1) return 0; | |
int max_profit = 0; | |
int buy_date = 0; | |
for(int i = 0; i < prices.length; i++){ | |
if(prices[i] < prices[buy_date]) buy_date = i; | |
int tmp_profit = prices[i] - prices[buy_date]; | |
if(tmp_profit > max_profit) max_profit = tmp_profit; | |
} | |
return max_profit; | |
} | |
} |
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/** | |
* Say you have an array for which the ith element is the price of a given stock on day i. | |
* Design an algorithm to find the maximum profit. | |
* You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). | |
* However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). | |
*/ | |
public class Solution{ | |
public int maxProfit(int[] prices) { | |
if(prices.length < 2) | |
return 0; | |
int max = 0; | |
for(int i = 1; i < prices.length; i++){ | |
int tmp_profit = prices[i] - prices[i-1]; | |
if(tmp_profit > 0) | |
max += tmp_profit; | |
} | |
return max; | |
} | |
} |
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/** | |
* Say you have an array for which the ith element is the price of a given stock on day i. | |
* Design an algorithm to find the maximum profit. You may complete at most two transactions. | |
* You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). | |
*/ | |
public class Solution{ | |
public int maxProfit(int[] prices) { | |
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; | |
int[] forward = new int[prices.length]; | |
int[] backward = new int[prices.length]; | |
for(int i = 0; i < prices.length; i++){ | |
// the first transaction profit | |
if(prices[i] > min) | |
forward[i] = Math.max(prices[i] - min, forward[i - 1]); | |
else | |
if(i > 0) | |
forward[i] = forward[i - 1]; | |
min = Math.min(prices[i], min); | |
// the second transaction | |
if(prices[prices.length - 1 - i] < max) | |
backward[prices.length - 1 - i] = Math.max(max - prices[prices.length - 1 - i], backward[prices.length - i]); | |
else | |
if(i > 0) | |
backward[prices.length - 1 - i] = backward[prices.length - i]; | |
max = Math.max(prices[prices.length - 1 - i], max); | |
} | |
// find out the max profit combined 2 transactions | |
int res = 0; | |
for(int i = 0; i < prices.length; i++) | |
res = Math.max(forward[i] + backward[i], res); | |
return res; | |
} | |
public int maxProfit(int[] P) { | |
if(P.length < 2) return 0; | |
int[] one = new int[P.length]; | |
int minP = P[0]; | |
for(int i = 1; i < P.length; i++) { | |
minP = Math.min(minP, P[i]); | |
one[i] = Math.max(one[i - 1], P[i] - minP); | |
} | |
int[] two = new int[P.length]; | |
int maxP = P[P.length - 1]; | |
for(int i = P.length - 2; i >= 0; i--) { | |
maxP = Math.max(maxP, P[i]); | |
two[i] = Math.max(two[i + 1], maxP - P[i]); | |
} | |
int res = Integer.MIN_VALUE; | |
for(int i = 0; i < P.length; i++) { | |
res = Math.max(one[i] + two[i], res); | |
} | |
return res; | |
} | |
public int maxProfit(int[] prices) { | |
int hold1 = Integer.MIN_VALUE, hold2 = Integer.MIN_VALUE; | |
int release1 = 0, release2 = 0; | |
for(int i:prices){ // Assume we only have 0 money at first | |
release2 = Math.max(release2, hold2+i); // The maximum if we've just sold 2nd stock so far. | |
hold2 = Math.max(hold2, release1-i); // The maximum if we've just buy 2nd stock so far. | |
release1 = Math.max(release1, hold1+i); // The maximum if we've just sold 1nd stock so far. | |
hold1 = Math.max(hold1, -i); // The maximum if we've just buy 1st stock so far. | |
} | |
return release2; ///Since release1 is initiated as 0, so release2 will always higher than release1. | |
} | |
} |
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