Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
- Input: [7, 1, 5, 3, 6, 4]
- Output: 5
- max difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
- Input: [7, 6, 4, 3, 1]
- Output: 0
- In this case, no transaction is done, i.e. max profit = 0.
C Solution 1:
int maxProfit(int* prices, int pricesSize) {
int res = 0, min = INT_MAX;
int i;
for (i = 0; i < pricesSize; i++) {
min = min < prices[i] ? min : prices[i];
res = res > prices[i] - min ? res : prices[i] - min;
}
return res;
}
C Solution 2:
int maxProfit(int* prices, int pricesSize) {
int res = 0, min = INT_MAX;
int i;
for (i = 0; i < pricesSize; i++) {
if (prices[i] <= min) {
min = prices[i];
continue;
}
res = res > prices[i] - min ? res : prices[i] - min;
}
return res;
}
Python Solution:
class Solution(object):
def maxProfit(self, prices):
"""
:type prices: List[int]
:rtype: int
"""
if not prices: return 0
minimum = prices[0]
res = 0
for price in prices[1:]:
res = max(res, price - minimum)
minimum = min(minimum, price)
return res
Summary:
- nothing to say.
LeetCode: 121. Best Time to Buy and Sell Stock
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