LeetCode(55) Jump Game

题目如下：

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.

分析如下：

从数组第0位开始，不断地向数组第n-1位扩展，每次扩展都记录当前能走到的最大距离，直到不能再走。

代码如下：

//http://www.cnblogs.com/TenosDoIt/p/3719630.html
// 10ms
class Solution {
public:
    bool canJump(int A[], int n) {
        int canArrive = 0;
        for(int i = 0; i <= canArrive && canArrive < n-1; i++)
            if(i + A[i] > canArrive)
                canArrive = i + A[i];
        return canArrive >= n-1;
    }
};

/*我的错误版本
class Solution {
public:
    bool canJump(int A[], int n) {
        if (n <= 0 ) return false;
        if (n == 1) return true;
        vector<int> B(n, 0);
        B[0] = 0;
        for (int i = 0; i < n; ++i) {
            if (B[i] + A[i] >= n) return true; 
            B[i] = i + A[i];
        }        
        return false;
    }
};

*/