From 8b35120ef45f817a6fc4f54ce7fe6a5978074d25 Mon Sep 17 00:00:00 2001
From: ShusenTang <tangshusen@pku.edu.cn>
Date: Fri, 3 Apr 2020 22:05:41 +0800
Subject: [PATCH] add 45. Jump Game II :beer:

---
 README.md                     |  1 +
 solutions/45. Jump Game II.md | 52 +++++++++++++++++++++++++++++++++++
 2 files changed, 53 insertions(+)
 create mode 100644 solutions/45. Jump Game II.md

diff --git a/README.md b/README.md
index f0c0d6c..a6db914 100644
--- a/README.md
+++ b/README.md
@@ -49,6 +49,7 @@ My LeetCode solutions with Chinese explanation. 我的LeetCode中文题解。
 | 42 |[Trapping Rain Water](https://leetcode.com/problems/trapping-rain-water/)|[C++](solutions/42.%20Trapping%20Rain%20Water.md)|Hard|  |
 | 43 |[Multiply Strings](https://leetcode.com/problems/multiply-strings/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/43.%20Multiply%20Strings.md)|Medium|  |
 | 44 |[Wildcard Matching](https://leetcode.com/problems/wildcard-matching/)|[C++](solutions/44.%20Wildcard%20Matching.md)|Hard|  |
+| 45 |[Jump Game II](https://leetcode.com/problems/jump-game-ii/) |[C++](solutions/45.%20Jump%20Game%20II.md)|Hard|  |
 | 46 |[Permutations](https://leetcode.com/problems/permutations/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/46.%20Permutations.md)|Medium|  |
 | 47 |[Permutations II](https://leetcode.com/problems/permutations-ii/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/47.%20Permutations%20II.md)|Medium|  |
 | 48 |[Rotate Image](https://leetcode.com/problems/rotate-image/)|[C++](https://github.com/ShusenTang/LeetCode/blob/master/solutions/48.%20Rotate%20Image.md)|Medium|  |
diff --git a/solutions/45. Jump Game II.md b/solutions/45. Jump Game II.md
new file mode 100644
index 0000000..ee0b338
--- /dev/null
+++ b/solutions/45. Jump Game II.md	
@@ -0,0 +1,52 @@
+# [45. Jump Game II](https://leetcode.com/problems/jump-game-ii/)
+
+# 思路
+
+跳格游戏，问最少跳多少步可跳到最后一个格子。
+
+因为我们需要用最少的步数跳到最后一个格子，所以可以考虑用贪心的思想，不过这里贪婪并不是每次都要跳到最远的格子，而是贪婪地求出用一定步数所能跳出最远的范围，一旦当这个范围到达末尾时，此时所用的步数一定是最小步数。所以基本思想是假设我们求出了用`step - 1`步能跳到的最远格子，那么就可以求出用`step`能跳到的最远范围。
+
+为此，先定义记录步数的变量`step`，再定义两个变量`cur`和`pre`分别来保存当前的能到达的最远位置和之前能到达的最远位置，然后从后往前遍历
+* 如果当前位置`i`小于等于pre，说明还是在`step`步能到达的范围内，根据当前位置格子的值来更新`cur`:`cur = max(cur, i + nums[i])`；
+* 如果当前位置`i`等于了pre，说明到达了`step`步能到达的范围内的最后一个格子，所以此时我们需要更新`pre`为`cur`，然后将step加1。
+
+时间复杂度O(n)，空间复杂度O(1)
+
+# C++
+## 写法一
+``` C++
+class Solution {
+public:
+    int jump(vector<int>& nums) {
+        int n = nums.size(), i = 0, cur = 0, pre = 0, step = 0;
+        for(int i = 0; i < n - 1; i++){
+            cur = max(cur, i + nums[i]);
+            if(i == pre){
+                pre = cur;
+                step++;
+                if(cur >= n - 1) return step;
+            }
+        }
+        return step;
+    }
+};
+```
+
+## 写法二
+``` C++
+class Solution {
+public:
+    int jump(vector<int>& nums) {
+        int n = nums.size(), i = 0, cur = 0, pre = 0, step = 0;
+        while(cur < n - 1){
+            while(cur < n - 1 && i <= pre)
+                cur = max(cur, i + nums[i++]);
+            pre = cur;
+            step++;
+        }
+        return step;
+    }
+};
+```
+
+