121. Best Time to Buy and Sell Stock
You are given an array prices where prices[i] is the price of a given stock on the ith day.
You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.
Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
Constraints:
- 1 <= prices.length <= 10^5
- 0 <= prices[i] <= 10^4
Example
Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
思路
分析题目
- 求两数之差最大值
- 只能高位减低位
那也就是说如果低位比高位大，那这个低位就应该被高位取代
题解
- 无脑快速AC
public int maxProfit(int[] prices) {
ArrayList<Integer> data = new ArrayList<>();
Arrays.stream(prices).forEach(data::add);
int left, right, result;
result = left = 0;
right = left + 1;
// 低位要小于高位且高位小于数组最大位数
while (left < right && right < prices.length) {
// 如果当前低位比高位大，那高位应该取代低位位置
if (prices[left] > prices[right]) {
left = right;
right = left + 1;
continue;
}
result = Math.max(result, prices[right] - prices[left]);
right++;
}
return result;
}
- 优化
public int maxProfit(int[] prices) {
ArrayList<Integer> data = new ArrayList<>();
Arrays.stream(prices).forEach(data::add);
int left, right, result;
result = left = 0;
right = left + 1;
while (right < prices.length) {
if (prices[left] > prices[right])
left = right;
else
result = Math.max(result, prices[right] - prices[left]);
right++;
}
return result;
}