Leetcode_best-time-to-buy-and-sell-stock-iii
地址：http://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete at most two transactions.
Note:
You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
2. 逆序遍历记录当前开始到最后一次可以获取的最大利润，记录并更新从尾开始的最大值。
如果在某个点有重复，即之前获得的最大值的终点和之后获得的最大值的起点是同一点，其实就是一次交易。比如 3 1 2 7 10， 顺序遍历会记录3->7，逆序会记录7->10，如果在这个点有最大值，其实就是一次交易。如果不发生在实际的同一点(注意当前交易所得并不一定是最大，可能会被之前更大的值更新)，那就是两次交易和。
参考代码：80ms
class Solution {
public:
int maxProfit(vector &prices) {
if(prices.size()<=1)
return 0;
int buy_low = INT_MAX, sell_high = INT_MIN, profit = 0, ans = 0;
vectorfirst_profit(prices.size(), 0);
for(int i = 0; i=0; --i)
{
if(prices[i]>sell_high)
sell_high = prices[i];
else
profit = max(profit, sell_high - prices[i]);
ans = max(ans, first_profit[i] + profit);
}
return ans;
}
};
//SECOND TRIAL, 60ms
class
Solution
{
public
:
int
maxProfit
(
vector
<
int
>
&
prices
)
{
if
(
prices
.
size
()
<=
1
)
return
0
;
vector
<
int
>
v1
(
prices
.
size
(),
0
),
v2
(
prices
.
size
(),
0
);
int
low
=
prices
[
0
],
high
=
prices
[
prices
.
size
()
-
1
];
for
(
int
i
=
1
;
i
<
prices
.
size
();
++
i
)
{
low
=
min
(
low
,
prices
[
i
]);
v1
[
i
]
=
max
(
v1
[
i
-
1
],
prices
[
i
]
-
low
);
}
for
(
int
j
=
prices
.
size
()
-
2
;
j
>=
0
;
--
j
)
{
high
=
max
(
high
,
prices
[
j
]);
v2
[
j
]
=
max
(
v2
[
j
+
1
],
high
-
prices
[
j
]);
}
int
ans
=
0
;
for
(
int
i
=
0
;
i
<
prices
.
size
();
++
i
)
ans
=
max
(
ans
,
v1
[
i
]
+
v2
[
i
]);
return
ans
;
}
};