题目:
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.
分析:
从后往前推算，keep track of maximum prices and maximum difference
public int maxProfit(int[] prices) {
if(prices.length == 0) return 0;
int max_diff = 0, max_number = prices[prices.length - 1];
for(int i = prices.length - 1; i >= 0; i--) {
max_diff = Math.max(max_diff, max_number - prices[i]);
max_number = Math.max(max_number, prices[i]);
}
return max_diff;
}
class Solution {
public int maxProfit(int[] prices) {
if(prices.length == 0) return 0;
int max_diff = 0;
int min = prices[0];
for(int i = 1; i < prices.length; i++) {
if(prices[i] - min > max_diff) max_diff = prices[i] - min;
if(prices[i] < min) min = prices[i];
}
return max_diff;
}
}
