1.假设股票只能买卖一次要求利润最大问题 https://leetcode.com/problems/best-time-to-buy-and-sell-stock/
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.
问题的关键点在于找出一个最小值,设为sofarMin,假设数组一个个就是sofarMin,最一直用后面元素比较即可
class Solution {
public int maxProfit(int[] prices) {
if(prices.length==0){
return 0;
}
int max = 0;//max初始值必须为0 一次买入卖出
int sofarMin = prices[0];
for(int i=1; i<prices.length; i++){
if(prices[i] > sofarMin){
max = Math.max(max,prices[i]-sofarMin);
}else{
sofarMin = prices[i];
}
}
return max;
}
}
2.假设股票可以连续买卖问题 https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/
Input: [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
class Solution {
public int maxProfit(int[] prices) {
int sum = 0;
for(int i=1; i<prices.length; i++){
if(prices[i]>prices[i-1]){
sum += (prices[i]-prices[i-1]);
}
}
return sum;
}
}