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121. 买卖股票的最佳时机
给定一个数组 prices ，它的第 i 个元素 prices[i] 表示一支给定股票第 i 天的价格。
你只能选择 某一天 买入这只股票，并选择在 未来的某一个不同的日子 卖出该股票。设计一个算法来计算你所能获取的最大利润。
返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0。
示例 1：
输入：[7,1,5,3,6,4]
输出：5
解释：在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。
注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。
示例 2：
输入：prices = [7,6,4,3,1] 输出：0 解释：在这种情况下, 没有交易完成, 所以最大利润为 0。
提示：
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1 <= prices.length <= 105 -
0 <= prices[i] <= 104
思路分析
一次遍历，在遍历中找最大值减去最小值即可。
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/**
* Runtime: 1 ms, faster than 69.16% of Java online submissions for Best Time to Buy and Sell Stock.
* Memory Usage: 42.6 MB, less than 5.31% of Java online submissions for Best Time to Buy and Sell Stock.
*
* Copy from: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/solution/yi-ge-fang-fa-tuan-mie-6-dao-gu-piao-wen-ti-by-lab/[一个方法团灭 6 道股票问题 - 最佳买卖股票时机含冷冻期 - 力扣（LeetCode）]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-01 13:01
*/
public int maxProfitDp(int[] prices) {
int dp0 = 0;
int dp1 = Integer.MIN_VALUE;
for (int i = 0; i < prices.length; i++) {
dp0 = Math.max(dp0, dp1 + prices[i]);
dp1 = Math.max(dp1, - prices[i]);
}
return dp0;
}
/**
* Runtime: 3 ms, faster than 19.73% of Java online submissions for Best Time to Buy and Sell Stock.
* Memory Usage: 42.7 MB, less than 5.31% of Java online submissions for Best Time to Buy and Sell Stock.
*
* Copy from: https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-with-cooldown/solution/yi-ge-fang-fa-tuan-mie-6-dao-gu-piao-wen-ti-by-lab/[一个方法团灭 6 道股票问题 - 最佳买卖股票时机含冷冻期 - 力扣（LeetCode）]
*/
public int maxProfitDpTable(int[] prices) {
if (Objects.isNull(prices) || prices.length == 0) {
return 0;
}
int length = prices.length;
int[][] dp = new int[length][2];
dp[0][0] = 0;
dp[0][1] = -prices[0];
for (int i = 1; i < length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + prices[i]);
dp[i][1] = Math.max(dp[i - 1][1], -prices[i]);
}
return dp[length - 1][0];
}
/**
* Runtime: 0 ms, faster than 100.00% of Java online submissions for Best Time to Buy and Sell Stock.
*
* Memory Usage: 37.3 MB, less than 100.00% of Java online submissions for Best Time to Buy and Sell Stock.
*/
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;
}
/**
* Runtime: 198 ms, faster than 8.08% of Java online submissions for Best Time to Buy and Sell Stock.
*
* Memory Usage: 38.5 MB, less than 76.99% of Java online submissions for Best Time to Buy and Sell Stock.
*/
public int maxProfitBruteForce(int[] prices) {
if (Objects.isNull(prices) || prices.length < 2) {
return 0;
}
int maxProfit = 0;
for (int i = 0; i < prices.length - 1; i++) {
for (int j = i + 1; j < prices.length; j++) {
int profit = prices[j] - prices[i];
if (profit > maxProfit) {
maxProfit = profit;
}
}
}
return maxProfit;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-15 07:44:20
*/
public int maxProfit(int[] prices) {
int result = 0;
int min = 100000;
for (int p : prices) {
min = Math.min(min, p);
result = Math.max(result, p - min);
}
return result;
}
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2025-04-10 20:48:34
*/
public int maxProfit(int[] prices) {
int min = Integer.MAX_VALUE;
int result = Integer.MIN_VALUE;
for (int price : prices) {
min = Math.min(min, price);
result = Math.max(result, price - min);
}
return result;
}