Jump Game II

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mob604756eba0ee 2014-11-23 18:35:00

文章标签 html ios 动态规划 c++ #include 文章分类 后端开发

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.

Your goal is to reach the last index in the minimum number of jumps.

For example:
Given array A = [2,3,1,1,4]

The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)

C++代码实现：

#include<iostream>
using namespace std;

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

int main()
{
    Solution s;
    int A[15]= {7,0,9,6,9,6,1,7,9,0,1,2,9,0,3};
    cout<<s.jump(A,15)<<endl;
}

使用动态规划的方法，一直超时。。。。。

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

另一种改进的方法：

ret:目前为止的jump数

curRch:从A[0]进行ret次jump之后达到的最大范围

curMax:从0~i这i+1个A元素中能达到的最大范围

当curRch < i，说明ret次jump已经不足以覆盖当前第i个元素，因此需要增加一次jump，使之达到

记录的curMax。

class Solution {
public:
    int jump(int A[], int n) {
        int ret = 0;
        int curMax = 0;
        int curRch = 0;
        for(int i = 0; i < n; i ++)
        {
            if(curRch < i)
            {
                ret ++;
                curRch = curMax;
            }
            curMax = max(curMax, A[i]+i);
        }
        return ret;
    }
};