提交 123 买卖股票的最佳时刻III

yllg · yllg · commit 8b5c65882c04 · 2019-06-05T15:03:25.000+08:00
diff --git a/123.买卖股票的最佳时机III/123 solution.html b/123.买卖股票的最佳时机III/123 solution.html
@@ -0,0 +1,62 @@
+<!DOCTYPE html>
+<html lang="en">
+<head>
+    <meta charset="UTF-8">
+    <meta name="viewport" content="width=device-width, initial-scale=1.0">
+    <meta http-equiv="X-UA-Compatible" content="ie=edge">
+    <title>Document</title>
+</head>
+<body>
+<script>
+// Source : https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
+// Author : 悬笔e绝
+// Date   : 2019-06-05
+
+/**
+ * @param {number[]} prices
+ * @return {number}
+ */
+ var maxProfit = function(prices) {
+  var len = prices.length
+    , sell = []
+    , buy = []
+    , item
+    , i;
+
+  if (!len)
+    return 0;
+
+  for (i = 0; i < len; i++) {
+    sell[i] = [],
+    buy[i] = [],
+    item = prices[i];
+    if (!i) {
+      sell[i][0] = 0;
+      sell[i][1]= 0;
+      sell[i][2] = 0;
+      buy[i][0] = -item;
+      buy[i][1] = -item;
+      buy[i][2] = -item;
+    } else {
+      sell[i][0] = sell[i - 1][0];
+      sell[i][1] = Math.max(sell[i - 1][1], buy[i - 1][1] + item);
+      sell[i][2] = Math.max(sell[i - 1][2], buy[i - 1][2] + item);
+      buy[i][0] = buy[i - 1][0];
+      buy[i][1] = Math.max(buy[i - 1][1], sell[i - 1][0] - item);
+      buy[i][2] = Math.max(buy[i - 1][2], sell[i - 1][1] - item);
+    }
+  }
+  return sell[len - 1][2];
+};
+
+// 测试
+var prices1 = [3,3,5,0,0,3,1,4];
+console.log(maxProfit(prices1));
+var prices2 = [1,2,3,4,5];
+console.log(maxProfit(prices2));
+var prices3 = [7,6,4,3,1];
+console.log(maxProfit(prices3));
+
+</script>
+</body>
+</html>
