Leetcode 55. Jump Game

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.

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思路：

思路1：
更新每一个点所能到的最大范围，当然这个点前提必须是可以到达的，如果最后都可以达到，返回true
思路2：
先定义最大距离为nums[0]，那么如果往后面的任何一个点所能走的距离大于这个，就把maxstep赋给它，让它走，如果某一个点能够到达的距离大于剩余距离了，说明可以走完

代码：

思路一代码：

class Solution {public:    bool canJump(vector<int>& nums) {        int size = nums.size();        int reach = 0;        int i = 0;        for(i = 0; i < size && i <= reach; ++i) {            reach = max(reach, nums[i]+i);        }        return i == size;    }};

思路二代码：

class Solution {public:    bool canJump(vector<int>& nums) {        int size=nums.size();        if(size<=0){            return false;        }        int maxJump=-1;        for(int i=0; i<size; i++){            if(nums[i]>maxJump){                maxJump=nums[i];            }            if(maxJump>=size-i-1){                return true;            }            if(maxJump==0){                return false;            }            maxJump--;        }        return false;    }};