1.一次买入一次卖出,O(n)算法
public class Solution {
public int maxProfit(int[] prices) {
int n = prices.length, minPrice = Integer.MAX_VALUE;
int res =0 ;
for(int i=0;i<n;i++){
if(prices[i]<minPrice)
minPrice = prices[i];
else if(prices[i]-minPrice > res){
res = prices[i]-minPrice;
}
}
return res;
}
}
2.多次买入,多次卖出,但是同时只能持有一个股票。此时只需要判断相邻两天是否递增即可
public class Solution {
public int maxProfit(int[] prices) {
int res = 0;
for(int i=1;i<prices.length;i++){
int tmp = prices[i]-prices[i-1];
if(tmp>0)
res += tmp;
}
return res;
}
}
3.两次买入两次卖出,不能同时持有
public class Solution {
public int maxProfit(int[] prices) {
if(prices.length==0) return 0;
int n = prices.length;
int buy1 = Integer.MIN_VALUE ,buy2 = Integer.MIN_VALUE , sell1 = 0, sell2 = 0;
for(int i=0;i<n;i++){
buy1 = Math.max(buy1,-prices[i]); //加负号最大实质上是最小
sell1 = Math.max(sell1,prices[i]+buy1);
buy2 = Math.max(buy2,sell1-prices[i]);
sell2 = Math.max(sell2,prices[i]+buy2);
}
return sell2;
}
}