LeetCode进阶之路（ Jump Game）

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.

题目：数组跳跃，按当前位置的值为最大跳跃距离，能否跳到最后一位。

思路：从倒数第二位往回遍历，找满足距离的那个位置，一直往下找，直到0；这种方法是方向的，感觉理解起来比较容易。

public boolean canJump(int[] nums) {        if(nums.length <= 1){            return true;        }        int last = nums.length-1;        for(int i = nums.length-2; i >=0 ;i--) {            if(i+nums[i] >= last) {                last = i;//            }                    }        return last<=0;    }

思路二：正向遍历

public boolean canJump(int[] nums) {        if(nums.length <= 1) {            return true;        }        int reach = 0;        for(int i = 0;i < nums.length - 1;i++) {            if(i + nums[i] > reach && i <= reach) {                reach = i + nums[i];            }        }        return reach >= nums.length-1;    }

种一棵树最好的时间是十年前，其次是现在！