Jump Game II

第一反应动态规划，代码如下，但是超时了

class Solution {
public:
    int jump(int A[], int n) {
        
        int* steps=new int[n];
        steps[0]=0;
        for(int i=1;i<n;++i)
        {
            int minStep=n*2;
            for(int j=0;j<i;++j)
            {
                if(((j+A[j])>=i)&&((steps[j]+1)<minStep))
                {
                    minStep=steps[j]+1;
                }
            }
            steps[i]=minStep;
        }
        return steps[n-1];
    }
};

上网查找后，找到了O(n)复杂度的算法，可以参考 http://www.cnblogs.com/lichen782/p/leetcode_Jump_Game_II.html

class Solution {
public:
    int jump(int A[], int n) {
        /*
 <span style="white-space:pre">	</span>* We use "last" to keep track of the maximum distance that has been reached
 <span style="white-space:pre">	</span>* by using the minimum steps "ret", whereas "curr" is the maximum distance
 <span style="white-space:pre">	</span>* that can be reached by using "ret+1" steps. Thus,
 <span style="white-space:pre">	</span>* curr = max(i+A[i]) where 0 <= i <= last.
 <span style="white-space:pre">	</span>*/
        int ret = 0;
        int last = 0;
        int curr = 0;
        for (int i = 0; i < n; ++i) {
            if (i > last) {
                last = curr;
                ++ret;
            }
            curr = max(curr, i+A[i]);
        }

        return ret;
    }
};