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MaxProfit.java
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80 lines (77 loc) · 2.67 KB
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package basic.dynamic;
/**
* leetcode121
* 买卖股票的最佳时间
*
* 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 (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
* Note that you cannot sell a stock before you buy one.
* <p>
* Example 1:
* Input: [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.
* Not 7-1 = 6, as selling price needs to be larger than buying price.
* Example 2:
* Input: [7,6,4,3,1]
* Output: 0
* Explanation: In this case, no transaction is done, i.e. max profit = 0.
*
* @author JunjunYang
* @date 2020/1/27 13:32
*/
public class MaxProfit {
public static void main(String[] args) {
System.out.println(new MaxProfit().maxProfit(new int[]{7, 1, 5, 3, 6, 4}));
}
/**
* 找到波谷，然后依次计算和波谷的差，得到最大利润
* @param prices
* @return
*/
public int maxProfit2(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;
}
/**
* 买卖股票
* 最多进行一次买卖
* 时间复杂度O(N)
* @param prices 每天的股票价格列表
* @return 买卖一次的最大利润，扩展：返回哪天买入哪天卖出
*/
public int maxProfit(int[] prices) {
//最终结果，最终结果对应的左右下标
int L = 0, R = 0, ans = 0;
//局部最大值,局部最大值对应的左右下标
int localMax = Integer.MIN_VALUE, localL = 0, localR = 0;
for (int i = 1; i < prices.length; i++) {
//如果之前累加的局部最大值小于0，则从当前节点开始重新累加
if (localMax < 0) {
localMax = prices[i] - prices[i - 1];
localL = i - 1;
localR = i;
} else {
localMax = localMax + prices[i] - prices[i - 1];
localR = i;
}
//取各个局部最大值中的最大值
if (localMax > ans) {
ans = localMax;
L = localL;
R = localR;
}
}
System.out.println(String.format("left index:%d, right index:%d, max profit:%d", L, R, ans));
return ans;
}
}