45. Jump Game II Leetcode Python

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本文介绍了一个算法问题：如何在给定非负整数数组中，从起始位置到达最后一个元素所需的最少跳跃次数。通过一次遍历整个数组并使用贪心策略实现，复杂度为O(n)。

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.)

This is also a greedy problem the difference between this one and the jump game is this time we need to count how many step we need to go from the start to the end.

we need to go through the whole array for once. The complexity if O(n).

we need two extra pointers "last" to track last element and count to track how many step we need to go.

code is as follow:

class Solution:
    # @param A, a list of integers
    # @return an integer
    def jump(self, A):
        index=0
        count=0
        last=0
        reach=0
        while reach>=index and index<len(A):
            if last<index:
                count+=1
                last=reach
            reach=max(reach,A[index]+index)
            index+=1
        if reach<len(A)-1:
            return 0
        else:
            return count