feat: 第 45 题添加新思路

Author: devin

docs/Leetcode_Solutions/Python/0045._Jump_Game_II.md

@@ -38,7 +38,7 @@ You can assume that you can always reach the last index.
 递归, 超时
 
 
-```
+```python
 class Solution:
     def jump(self, nums):
         """
@@ -67,7 +67,7 @@ DP
 
 dp[i]代表的是到达index为idx的位置的最少步数, 依然超时，最后2个case过不了
 
-```
+```python
 class Solution:
     def jump(self, nums):
         """
@@ -120,3 +120,27 @@ class Solution:
 ```
 
 
+> 思路 4
+
+要想获得最小跳跃次数，每次在可跳范围内选择势能最高的点，即下次能跳得最远的点，是：num[i]+i最大值，而不是 num[i]
+
+```python
+class Solution:
+    def jump(self, nums: List[int]) -> int:        
+        if len(nums) == 1:  # 只有一个元素的情况，直接返回 0
+            return 0
+        step = 1
+        i = 0
+        while i < len(nums)-1:
+            if i + nums[i] >= len(nums)-1:  # 当前可达，直接返回
+                return step
+            else:
+                max_num = nums[i+1]+i
+                max_j = i+1
+                for j in range(i+1, i+nums[i]+1):  # 在可跳范围内找下次能跳最远的点
+                    if nums[j]+j >= max_num:
+                        max_num = nums[j]+j
+                        max_j = j
+            step += 1
+            i = max_j
+```