[Lintcode]116. Jump Game/[Leetcode]55. Jump Game

116. Jump Game/55. Jump Game

• 本题难度: Medium
• Topic: Greedy

Description

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
A = [2,3,1,1,4], return true.

A = [3,2,1,0,4], return false.

Notice
This problem have two method which is Greedy and Dynamic Programming.

The time complexity of Greedy method is O(n).

The time complexity of Dynamic Programming method is O(n^2).

We manually set the small data set to allow you pass the test in both ways. This is just to let you learn how to use this problem in dynamic programming ways. If you finish it in dynamic programming ways, you can try greedy method to make it accept again.

我的代码

class Solution:
    """
    @param A: A list of integers
    @return: A boolean
    """
    def canJump(self, A):
        # write your code here
        l = len(A)-1
        maxpos = 0
        for i in range(l+1):
            print(i,maxpos,i+A[i])
            if maxpos < i:
                return False
            if maxpos < i + A[i]:
                maxpos = i + A[i]
                print('new',i, maxpos)
                if maxpos >= l:
                    return True
        return True
                

思路

最长能走到的位置和当前位置做比较

• 时间复杂度 O(n)
• 出错： 忘了A长度为1的情况。注意边界条件，好像这道题leetcode比lintcode严格。