方法一:(自己写的)
class Solution {
public int maxProfit(int[] prices) {
if(prices.length==0 || prices.length==1) {
return 0;
}
int p=0,q=1,profit=0;
while(q<prices.length) {
if(prices[p]>prices[q]) {
p=q;
q++;
}else {
while(q<prices.length-1 && prices[q+1]>prices[q]) {
q++;
}
profit+=prices[q]-prices[p];
p=q+1;
q=p+1;
}
}
return profit;
}
}
方法二:
对于 [a, b, c, d],如果有 a <= b <= c <= d ,那么最大收益为 d - a。而 d - a = (d - c) + (c - b) + (b - a) ,因此当访问到一个 prices[i] 且 prices[i] - prices[i-1] > 0,那么就把 prices[i] - prices[i-1] 添加到收益中。
class Solution {
public int maxProfit(int[] prices) {
int profit=0;
for(int i=1;i<prices.length;i++) {
if(prices[i]>prices[i-1]) {
profit+=prices[i]-prices[i-1];
}
}
return profit;
}
}