From 028efb9e7faa09b83009eb902bfd2b46c5e2be49 Mon Sep 17 00:00:00 2001
From: yi-ge <a@wyr.me>
Date: Wed, 6 May 2020 13:52:17 +0800
Subject: [PATCH] =?UTF-8?q?add:=20=E6=9C=80=E4=BD=8E=E7=A5=A8=E4=BB=B7?=
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---
 .vscode/settings.json                       |  1 +
 README.md                                   |  6 ++-
 src/array/minimum-cost-for-tickets.js       | 43 +++++++++++++++++++++
 test/array/minimum-cost-for-tickets.test.js |  6 +++
 4 files changed, 55 insertions(+), 1 deletion(-)
 create mode 100644 src/array/minimum-cost-for-tickets.js
 create mode 100644 test/array/minimum-cost-for-tickets.test.js

diff --git a/.vscode/settings.json b/.vscode/settings.json
index 38e1d7b..29cc9c7 100644
--- a/.vscode/settings.json
+++ b/.vscode/settings.json
@@ -3,6 +3,7 @@
     "dvdf",
     "lcci",
     "lcof",
+    "mincost",
     "nums",
     "pwwkew",
     "zhong"
diff --git a/README.md b/README.md
index 07c1b10..d0bce6a 100644
--- a/README.md
+++ b/README.md
@@ -253,7 +253,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
 - [最大数和最小数](src/array/maximum-and-minimum.js)
 
-  LintCode 770. 最大数和最小数 https://www.lintcode.com/problem/maximum-and-minimum/description
+  - LintCode 770. 最大数和最小数 https://www.lintcode.com/problem/maximum-and-minimum/description
+
+- [最低票价](src/array/minimum-cost-for-tickets.js)
+
+  - LeetCode 983. 最低票价 https://leetcode-cn.com/problems/minimum-cost-for-tickets/
 
 ## 栈
 
diff --git a/src/array/minimum-cost-for-tickets.js b/src/array/minimum-cost-for-tickets.js
new file mode 100644
index 0000000..ab2d320
--- /dev/null
+++ b/src/array/minimum-cost-for-tickets.js
@@ -0,0 +1,43 @@
+/**
+ * @param {number[]} days
+ * @param {number[]} costs
+ * @return {number}
+ */
+export const mincostTickets = function (days, costs) {
+  //   动态规划:
+  //   dp[i]: 从第i天开始，到最后一天所用的票价总和
+
+  //   思路：
+  //   1. 记录数组中的最早的一天和最后的一天，动态规划从最后一天向前到最早一天即可，因为
+  //      其他时间不需要消耗通行证
+  //   2. 用一个变量 k 指针从 days 的最后一个索引向前走，i 指针也从最后一天向前走，但是遇到了
+  //      不需要花费通行证的某些天走 else 分支即可，即后一天的总花费就是这一天的总花费
+  //   3. 如果遇到需要通行证的时候，分别对比买一天、七天、三十天、所需的总花费，选择总花费最少的策略即可
+  //   4. 初始化的时候 dp 数组多了 30，是为了简化有的用例第 365 天需要出行，选择买 30 天
+  //      通行证的时候的特判
+
+  // 作者：ignore_express
+  // 链接：https://leetcode-cn.com/problems/minimum-cost-for-tickets/solution/js-dong-tai-gui-hua-si-lu-jiang-jie-by-ignore_expr/
+  // 来源：力扣（LeetCode）
+  // 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
+  const dp = new Array(366 + 30).fill(0)
+  const n = days.length
+  const maxDay = days[n - 1]
+  const minDay = days[0]
+  let k = n - 1
+
+  for (let i = maxDay; i >= minDay; i--) {
+    if (i === days[k]) {
+      dp[i] = Math.min(
+        dp[i + 1] + costs[0],
+        dp[i + 7] + costs[1],
+        dp[i + 30] + costs[2]
+      )
+      k--
+    } else {
+      dp[i] = dp[i + 1]
+    }
+  }
+
+  return dp[minDay]
+}
diff --git a/test/array/minimum-cost-for-tickets.test.js b/test/array/minimum-cost-for-tickets.test.js
new file mode 100644
index 0000000..97ed54a
--- /dev/null
+++ b/test/array/minimum-cost-for-tickets.test.js
@@ -0,0 +1,6 @@
+import { mincostTickets } from '../../src/array/minimum-cost-for-tickets'
+
+test('最低票价', () => {
+  expect(mincostTickets([1, 4, 6, 7, 8, 20], [2, 7, 15])).toBe(11)
+  expect(mincostTickets([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 30, 31], [2, 7, 15])).toBe(17)
+})