算法分析與設計課程（9）：【leetcode】Jump Game

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.

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

A = [3,2,1,0,4], return false.

算法分析：

Jump Game是一道有意思的題目。題意很簡單，給你一個數組，數組的每個元素表示你能前進的最大步數，最開始時你在第一個元素所在的位置，之後你可以前進，問能不能到達最後一個元素位置。

我們可以考慮用動態規劃的方法解決

代碼如下：

class Solution:
    # @param A, a list of integers
    # @return an integer
    # def jump(self, A):
    #     maxint = 1<<31 - 1
    #     dp = [ maxint for i in range(len(A)) ]
    #     dp[0] = 0
    #     for i in range(1, len(A)):
    #         for j in range(i):
    #             if A[j] >= (i - j):
    #                 dp[i] = min(dp[i], dp[j] + 1)
    #     return dp[len(A) - 1]
    # dp is time limited exceeded!
    
# We use "last" to keep track of the maximum distance that has been reached
# by using the minimum steps "ret", whereas "curr" is the maximum distance
# that can be reached by using "ret+1" steps. Thus,curr = max(i+A[i]) where 0 <= i <= last.
    def jump(self, A):    
        ret = 0
        last = 0
        curr = 0
        for i in range(len(A)):
            if i > last:
                last = curr
                ret += 1
            curr = max(curr, i+A[i])
        return ret