上升子序列最大差(dp经典)

Author: cls1991

DynamicProgramming/121_BestTimeToBuyAndSellStock.py

@@ -0,0 +1,50 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/best-time-to-buy-and-sell-stock/description.
+    题目描述:
+
+    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.
+
+"""
+
+
+class Solution(object):
+    def maxProfit(self, prices):
+        """
+        :type prices: List[int]
+        :rtype: int
+        """
+        if not prices:
+            return 0
+
+        return self.dynamic_max_profit(prices)
+
+    # 动态规划
+    # 搜索局部上升子序列, 保存当前最优解
+    def dynamic_max_profit(self, prices):
+        start = 0
+        ans = 0
+        for i in range(1, len(prices)):
+            delta = prices[i] - prices[start]
+            if delta > 0:
+                ans = max(ans, delta)
+            else:
+                start = i
+
+        return ans

DynamicProgramming/53_MaximumSubarray.py

@@ -36,8 +36,7 @@ def dynamic_sub_array(self, nums):
         cur = 0
         for i in range(len(nums)):
             cur += nums[i]
-            if cur > ans:
-                ans = cur
+            ans = max(ans, cur)
             if cur < 0:
                 cur = 0