Leetcode 45. Jump Game II

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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.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

Note:
You can assume that you can always reach the last index.

解题思路：

检查第n个jump最远能到哪里，如果第n步的maxReach >= n-1, 那么答案就是n。第一跳的maxReach就是numns[0] (图例中等于5), 第二跳的maxReach是有head1到max1决定的。同理第三跳的maxReach是有head2和max2决定的。以此类推

 1 class Solution(object):
 2     def jump(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: int
 6         """
 7         n = len(nums)
 8         njumps = maxReach = head = 0
 9         
10         while maxReach < n-1:
11             njumps += 1
12             curMax = maxReach
13             for i in range(head, maxReach+1):
14                 curMax = max(curMax, i+nums[i])
15             head = maxReach + 1
16             maxReach = curMax
17         
18         return njumps

Leetcode 45. Jump Game II

标签：osi   div   sum   rom   repr   技术分享   for   init   nim   

原文地址：http://www.cnblogs.com/lettuan/p/6258126.html