121. Best Time to Buy and Sell Stock
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
同53题.
The maximum (minimum) subarray problem. 用到(DP).
这题是重点:
就是序列从A[0]开始计算, 不符合条件就清0, 从断处继续计算. 好处是 (O(n)) time 就可以处理, 牛逼的不行不行的.
(O(n)) time, (O(1)) space.
照葫芦画瓢代码:
//是个 最大子序列问题 相对应的 有最小子序列问题
//Kadane’s algorithm uses the dynamic programming approach
//to find the maximum (minimum) subarray
int maxProfit(vector<int>& A) {
int i = 1, profit = 0, temp = 0, p = 0;
while (i < A.size()) {
temp = A[i] - A[i - 1];
p = max(0, p += temp); //难点在这,p要和temp累积,小于0则清0
profit = max(p, profit); //profit记录累积过程中的最大收益值
i++;
}
return profit;
}
int max(int a, int b) {
return a > b ? a : b;
}
更牛逼的代码:
//TODO 还没看呢
int maxProfit(vector<int>& prices) {
int mini = INT_MAX;
int pro = 0;
for (int i = 0; i < prices.size(); i++) {
mini = min(prices[i], mini);
pro = max(pro, prices[i] - mini);
}
return pro;
}