- leetcode:121. Best Time to Buy a
- LeetCode #121 #122 #123 #188 #30
- #121. Best Time to Buy and Sell
- [LeetCode] 121. Best Time to Buy
- Leetcode121 - 123 （dp problems）
- [数组]121. Best Time to Buy and Se
- 121. Best Time to Buy and Sell S
- Leetcode121-Best Time to Buy and
- 每天(?)一道LeetCode(14) Best Time to
- Leetcode 股票合集 【121、122、714、309】
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)
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.
Already Pass Solution
public int MaxProfit(int[] prices) {
int result = 0;
int curMax = 0;
for(int i = 1; i < prices.Length; i++)
{
curMax += prices[i] - prices[i - 1];
curMax = Math.Max(0, curMax);
result = Math.Max(curMax,result);
}
return result;
}
解题思路：
1.使用结合律获取最大差值
待思考：
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