Dynamic Programming

 参考：http://fisherlei.blogspot.com/2012/12/leetcode-jump-game.html

javascript:void(0)

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.

 

C++实现代码：

#include<iostream>
using namespace std;

class Solution
{
public:
    bool canJump(int A[],int n)
    {
        int maxSum=0;
        int sum=0;
        int i;
        for(i=0;i<=maxSum&&i<n;i++)
        {
            sum=i+A[i];
            if(sum>maxSum)
                maxSum=sum;
            if(maxSum>=n-1)
                return true;
        }
        return false;
    }
};

int main()
{
    Solution s;
    int A[5]={3,2,1,0,4};
    cout<<s.canJump(A,5)<<endl;
}

 重新刷题，只能想到的笨办法。。。

class Solution {
public:
    bool canJump(int A[], int n) {
        if(n<=0)
            return false;
        int dp[n];
        memset(dp,false,sizeof(dp));
        dp[0]=true;
        int i,j;
        for(i=1;i<n;i++)
        {
            j=i-1;
            while(j>=0)
            {
                if(dp[j]&&j+A[j]>=i)
                {
                    dp[i]=true;
                    break;
                }
                --j;
            }
        }
        return dp[n-1];
    }
};