LeetCode--No.121--Best Time to Buy and Sell Stock

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.

方法：

时间复杂度O(n^2)的就略过了，比较简单。

方法二：

时间复杂度O(n),空间复杂度O(n), 其实和求子数列的最大和，几乎是一样的。

public class Solution {
public int maxProfit(int[] prices) {
int max = 0;
int min = 0;
if (prices.length <= 1)
return 0;
else if(prices.length == 2){
if (prices[1] - prices[0] > 0)
return prices[1] - prices[0];
else
return 0;
}
int[] sum = new int[prices.length - 1];
sum[0] = prices[1] - prices[0];
int maxs = sum[0];
for(int i = 1; i < prices.length-1; i++){
sum
4000
[i] = Math.max(prices[i+1] - prices[i], sum[i-1] + prices[i+1] - prices[i]);
maxs = Math.max(maxs, sum[i]);
}
if (maxs < 0)
return 0;
else
return maxs;
}
}

为什么就不能简单一点：？！！
感觉受到了一万点伤害。
我这个智商真的还要继续刷下去么？
public class Solution {
public int maxProfit(int[] prices) {
if (prices.length == 0)
return 0;
int maxPrice = prices[prices.length-1];
int ans = 0;
for(int i = prices.length - 1; i>= 0; i--){
maxPrice = Math.max(maxPrice, prices[i]);
ans = Math.max(ans, maxPrice - prices[i]);
}
return ans;
}
}