From b806ec7959f2b4de0df46d514de2caf0e39ac22e Mon Sep 17 00:00:00 2001
From: yige <a@wyr.me>
Date: Mon, 4 May 2020 13:43:56 +0800
Subject: [PATCH] =?UTF-8?q?add:=20=E8=B7=B3=E8=B7=83=E6=B8=B8=E6=88=8F=20I?=
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---
 README.md                       |  5 +++++
 src/array/jump-game-ii.js       | 20 ++++++++++++++++++++
 test/array/jump-game-ii.test.js |  5 +++++
 3 files changed, 30 insertions(+)
 create mode 100644 src/array/jump-game-ii.js
 create mode 100644 test/array/jump-game-ii.test.js

diff --git a/README.md b/README.md
index f5d1616..a0036ef 100644
--- a/README.md
+++ b/README.md
@@ -235,6 +235,11 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 面试题42. 连续子数组的最大和 https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/
   - LintCode 41. 最大子数组 https://www.lintcode.com/problem/maximum-subarray/description
 
+- [跳跃游戏II](src/array/jump-game-ii.js)
+
+  - LeetCode 45. 跳跃游戏 II https://leetcode-cn.com/problems/jump-game-ii/
+  - LintCode 117. 跳跃游戏 II https://www.lintcode.com/problem/jump-game-ii/description
+
 ## 栈
 
 - [最大矩阵](src/stack/maximal-rectangle.js)
diff --git a/src/array/jump-game-ii.js b/src/array/jump-game-ii.js
new file mode 100644
index 0000000..ce9cf76
--- /dev/null
+++ b/src/array/jump-game-ii.js
@@ -0,0 +1,20 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+export const jump = function (nums) {
+  let step = 0 // 跳跃步数
+  let maxPosition = 0 // 最大位置
+  let lastJumpStepMax = 0 // 最后一次跳跃最大能跳的步数
+
+  for (let n = 0, len = nums.length; n < len - 1; n++) {
+    maxPosition = Math.max(maxPosition, nums[n] + n)
+    if (lastJumpStepMax === n) { // 从当前位置开始，到当前能跳的位置这个区间，谁最大，就选谁，并跳一次
+      lastJumpStepMax = maxPosition
+      step++
+    }
+    if (lastJumpStepMax > len) return step
+  }
+
+  return step
+}
diff --git a/test/array/jump-game-ii.test.js b/test/array/jump-game-ii.test.js
new file mode 100644
index 0000000..8d44d96
--- /dev/null
+++ b/test/array/jump-game-ii.test.js
@@ -0,0 +1,5 @@
+import { jump } from '../../src/array/jump-game-ii'
+
+test('跳跃游戏II', () => {
+  expect(jump([2, 3, 1, 1, 4])).toBe(2)
+})