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Leetcode-Jump Game

时间:2014-11-17 08:03:12      阅读:181      评论:0      收藏:0      [点我收藏+]

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.

Have you met this question in a real interview?
 
Analysis:
We record the max reachable length and update it at each A[i].
1. If maxReach<i, then cannot reach i return false;
2. If maxReach<i+A[i], then we update the maxReach to i+A[i].
3. Once maxReach>=A.length-1, we can reach the end, return true.
 
Solution:
 1 public class Solution {
 2     public boolean canJump(int[] A) {
 3         if (A.length==0 || A.length==1) return true;
 4         int len = A.length;
 5 
 6         int maxReach = A[0];
 7         for (int i=1;i<len;i++)
 8             if (maxReach<i)
 9                 return false;
10             else {
11                 if (maxReach<i+A[i]) maxReach = i+A[i];
12                 if (maxReach>=len-1) return true;
13             }
14        
15         return false;       
16     }
17 }

 

Leetcode-Jump Game

原文:http://www.cnblogs.com/lishiblog/p/4102809.html

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