[leetcode] 123. 买卖股票的最佳时机 III
暴力做法
直接枚举中间值,分成两个121. 买卖股票的最佳时机题做:
class Solution {
public int maxProfitOne(int[] prices) {
int min = Integer.MAX_VALUE;
int ans = 0;
for (int price : prices) {
if (price < min) {
min = price;
} else if (ans < price - min) {
ans = price - min;
}
}
return ans;
}
public int maxProfit(int[] prices) {
int ans = 0;
for (int i = 0; i < prices.length; i++) {
int[] left = new int[i];
int[] right = new int[prices.length - i];
System.arraycopy(prices, 0, left, 0, i);
System.arraycopy(prices, i, right, 0, prices.length - i);
ans = Math.max(ans, maxProfitOne(left) + maxProfitOne(right));
}
return ans;
}
}