题目：

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.)

思路：

第一种思路是利用迭代的思路来计算最小跳数，但是时间复杂度比较大；第二种思路是反过来想，要达到最后一条，倒数第二条至少应该到哪个位置，以此类推直到我们倒推到第一位时便可知最小跳数；第三种思路是用动态规划DP的观点来实现。DP[i]代表到达i的最小跳数，显然DP是一个递增的数组。每次循环只需要尽量找到最小的DP[k]，使其满足k+A[k]>=n。

代码：

思路一：迭代（时间复杂度不满足要求）

class Solution {
public:
    int jump(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int* v = new int[1];
        v[0] = INT_MAX;
        
        jumpRepeat(A, 0, n-1, 0, v);
        
        if(v[0] == INT_MAX)
        {
            return 0;
        }
        
        return v[0];
    }
    
    void jumpRepeat(int A[], int i, int m, int n,int* v)
    {
        if(i >= m)
        {
            if(v[0] > n)
            {
                v[0] = n;
            }
            return;
        }
        if(A[i] == 0)
        {
            return;
        }
        else
        {
            for(int j = 1; j <= A[i]; j++)
            {
                jumpRepeat(A, i + j, m, n+1, v);
            }
        }
    }
};

思路二：倒推

class Solution {
public:
    int jump(int A[], int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int pre = 0;
        int cur = n - 1;
        int count = 0;
        while(true)
        {
            if(pre == cur)
            {
                return 0;
            }
            count++;
            pre = cur;
            for(int i = n - 2; i >= 0; i--)
            {
                if(i + A[i] >= pre)
                {
                    if(cur > i)
                    {
                        cur = i;
                    }
                }
            }
            if(cur == 0)
            {
                return count;
            }
        };
        
   
    }
};

思路三：动态规划

class Solution {
public:
    int* dp;
    int jump(int A[], int n) {
        if(n==0)
        {
            return INT_MAX;
        }
        dp = new int[n];
        dp[0] = 0;
        for(int i=1;i<n;i++)
        {
            dp[i] = INT_MAX;
        }
        for(int i=1;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(j+A[j]>=i)
                {
                    int tmp = dp[j]+1;
                    if(tmp < dp[i])
                    {
                        dp[i] = tmp;
                        break;
                    }
                }
            }
        }
        
        return dp[n-1];
    }
};