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题目：
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.
思路：这样想，只要是正数，就能一直往前走，唯一障碍就是0，只要能跳跃过0就行了。所以每当我们遇到0时，就看之前的最大步数能不能跳过它。
package dp; public class JumpGame { public boolean canJump(int[] nums) { int len = nums.length; int max = 0; for (int i = 0; i < len - 1; ++i) { if (i + nums[i] > max) max = i + nums[i]; if (nums[i] == 0 && i >= max) // 为0，之前的最大步数不能跳过它就返回false return false; } return max >= len - 1; } public static void main(String[] args) { // TODO Auto-generated method stub int[] nums1 = {2,3,1,1,4}; int[] nums2 = {0,3,2}; JumpGame j = new JumpGame(); System.out.println(j.canJump(nums1)); System.out.println(j.canJump(nums2)); } }
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5013292.html
