leetcode.55.Jump Game(medium)[贪心算法]

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.

class Solution {
public:
    int max(int a,int b)
    {
        return a>=b?a:b;
    }
    bool canJump(vector<int>& nums) {
       /* int index = 0;
        int n = nums.size();
        while(index < n-1 && nums[index] != 0 )
        {
             index += nums[index];
        }
        if(index == n-1)
            return true;
        return false;*/
        //采用贪心算法，题目的每一步给出的当前位置能迈出的最大的步数，因此每次都比较上一步得到的最大位置和当前新的位置+可迈步数，取最大的作为可迈
        if(nums.size() == 0) return false;
        int start = 0;
        for(int i=0;i<=start&&i<nums.size();i++)
        {
            start = max(nums[i]+i,start);
        }
        if(start>=nums.size()-1)
           return true;
        return false;
    }
};