121. Best Time to Buy and Sell Stock - jiejackyzhang/leetcode-note GitHub Wiki
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.
Example 1:
Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1] Output: 0 In this case, no transaction is done, i.e. max profit = 0.
解题思路为Dynamic Programming。 维持两个变量minPrice和maxProfit。 顺序扫描数组，若prices[i]<minPrice，则更新minPrice；若prices[i]-minPrice>maxProfit，更新maxProfit。
public class Solution {
public int maxProfit(int[] prices) {
int minPrice = Integer.MAX_VALUE;
int maxProfit = 0;
for(int i = 0; i < prices.length; i++) {
if(prices[i] < minPrice) {
minPrice = prices[i];
} else if(prices[i] - minPrice > maxProfit) {
maxProfit = prices[i] - minPrice;
}
}
return maxProfit;
}
}more concise version:
public class Solution {
public int maxProfit(int[] prices) {
int res = 0;
int buy = Integer.MAX_VALUE;
for(int i : prices) {
buy = Math.min(buy, i);
res = Math.max(res, i-buy);
}
return res;
}
}