LeetCode 55. Jump Game

https://leetcode.com/problems/jump-game/

Medium

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.

---

• DP / Greedy. 
• 動規。定義f[i]爲i位置最大能jump到的index，f[i] = max{ f[i - 1], i + nums[i] }, f[0] = nums[0]，f[n]爲所求結果。中間注意能夠剪枝。
• 貪心。從後往前倒推，若是當前位置能跳到lastpos，則更新lastpos爲i。最後判斷lastpos是否爲0。如果則表明能從頭跳到尾。
• https://leetcode.com/problems/jump-game/solution/
• https://leetcode.com/problems/jump-game/discuss/114802/Python-DP-O(n)-solution-accepted

 1 class Solution:
 2     def canJump1(self, nums: List[int]) -> bool:
 3         lastpos = len(nums) - 1
 4         
 5         for i in range(len(nums) - 1, -1, -1):
 6             if i + nums[i] >= lastpos:
 7                 lastpos = i
 8         
 9         return lastpos == 0
10 
11     def canJump(self, nums: List[int]) -> bool:
12         f = [0] * len(nums)
13         f[0] = nums[0]
14         
15         for i in range(1, len(nums)):
16             if f[i - 1] < i:
17                 return False
18             
19             f[i] = max(f[i - 1], i + nums[i])
20         
21         return f[-1] >= len(nums) - 1

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