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LeetCode121.java
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105 lines (93 loc) · 3 KB
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package problems;
import java.util.Scanner;
public class LeetCode121 {
// 常规方法
public static int maxProfit(int[] prices) {
int minNum = 9999;
int ans = 0;
for(int i=0; i<prices.length; i++){
minNum = Math.min(minNum,prices[i]);
ans = Math.max(ans, prices[i]-minNum);
}
return ans;
}
// 动态规划解法
public static int maxProfitDP(int[] prices) {
int length=prices.length;
if(length<2){
return 0;
}
int [][] dp =new int [length][2];
// dp[i][0] 下标为 i 这天结束的时候，不持股，手上拥有的现金数
// dp[i][1] 下标为 i 这天结束的时候，持股，手上拥有的现金数
// 初始化：不持股显然为0，持股就需要减去第1天（下标为0）的股价
dp[0][0]=0;
dp[0][1]=-prices[0];
// 从第二天开始遍历
/*
* dp[i][0]有两种情况：
* dp[i-1][1] 前一天持股，今天卖出。
* dp[i-1][0] 前一天不持股，保持不变。
*
* dp[i][1]也有两种情况：
* dp[i-1][1] 前一天持股，今天扔持股。
* dp[i-1][0] 前一天不持股，今天买入。
*/
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];
}
public int maxProfit1(int[] prices) {
int max = 0;
int[] dp = new int[prices.length];
dp[0] = prices[0];
for (int i = 1; i < prices.length; i++) {
dp[i] = Math.min(dp[i - 1], prices[i]);
max = Math.max((prices[i] - dp[i]), max);
}
return max;
}
public static void main(String[] args) {
Scanner scanner=new Scanner(System.in);
int n = scanner.nextInt();
int[] prices = new int[n];
for(int i=0;i<prices.length;i++){
prices[i]=scanner.nextInt();
}
int ans = maxProfitDP(prices);
System.out.println(ans);
}
}
class LeetCode121_1{
public static int maxProfit(int[] prices) {
int n = prices.length;
int[] dp = new int[n];
int minCost = prices[0];
int maxProfit = 0;
for (int i = 1; i < n; i++) {
if(prices[i] < minCost){
minCost = prices[i];
}
dp[i] = Math.max(dp[i-1], prices[i] - minCost);
maxProfit = Math.max(maxProfit, dp[i]);
}
return maxProfit;
}
public static void main(String[] args) {
maxProfit(new int[]{1,2});
}
}
class LeetCode121_2{
public int maxProfit(int[] prices) {
int n = prices.length;
int minCost = prices[0];
int maxProfit = 0;
for (int i = 1; i < n; i++) {
minCost = Math.min(minCost, prices[i]);
maxProfit = Math.max(maxProfit, prices[i] - minCost);
}
return maxProfit;
}
}