Create 045._Jump_Game_II.md

045._Jump_Game_II.md

+### 45. Jump Game II
+
+题目:
+<https://leetcode.com/problems/jump-game-ii/>
+
+
+难度:
+
+Easy
+
+
+思路
+
+# greedy solution, the current jump is ```[i, cur_end]```, and the ```cur_farthest``` is the farthest point 
+# that all of point in ```[i, cur_end]``` can reach, whenever ```cur_farthest``` is larger than the last point' index, 
+# return current ```jump+1```; whenever ```i``` reaches ```cur_end```, update ```cur_end``` to ```current cur_farthest```.
+# Time: O(log(n))
+# Space: O(1)
+
+```python
+class Solution(object):
+    def jump(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        # Note You can assume that you can always reach the last index.
+        cur_end, cur_farthest, step, n = 0, 0, 0, len(nums)
+        for i in range(n-1):
+            cur_farthest = max(cur_farthest, i + nums[i])
+            if cur_farthest >= n - 1:
+                step += 1
+                break
+            if i == cur_end:
+                cur_end = cur_farthest
+                step += 1
+        return step
+
+            
+        
+```
+
+